A continuous non-degenerate symmetric tensor field on a connected manifold has the same signature everywhere. This is kind of like the intermediate-value theorem for real-valued functions: if a function is continuous, then the only way it can change sign (read: change signature) is if it has a root somewhere in between (read: becomes degenerate somewhere).
Before we prove this, we need a lemma at the level of vector spaces (I'm going to assume you know facts like every finite-dimensional real vector space has a norm, and all such norms are equivalent (i.e generate the same topology), that all multilinear maps are continuous, and that the Heine-Borel theorem holds in this case as well, etc).
Lemma.
Let $E$ be a real vector space of finite dimension $n$, let $\mathcal{S}(E)$ be the space of symmetric $(0,2)$ tensors on $E$. For any pair $(p,q)$ of non-negative integers such that $p+q=n$ let us define $\mathcal{S}_{p,q}(E)$ to be the set of symmetric $(0,2)$ tensors on $E$ with signature $(p,q)$ (plus, minus). Then, $\mathcal{S}_{p,q}(E)$ is open in $\mathcal{S}(E)$.
To prove this, fix a tensor $T_0\in\mathcal{S}_{p,q}(E)$. By hypothesis, there is a direct sum decomposition $E=P\oplus N$ into "positive" and "negative" subspaces of $T_0$, i.e for all $x\in P\setminus\{0\}$ we have $T_0(x,x)>0$ and for all $x\in N\setminus\{0\}$, we have $T_0(x,x)<0$. Fix any norm $\|\cdot\|$ on $E$.
- The function $x\mapsto T_0(x,x)$ defined on the unit sphere of $P$, $\Bbb{S}_P:=\{x\in P\,:\, \|x\|=1\}$ is continuous and positive, so by compactness of the unit sphere, it has a positive infimum $\beta_+>0$. By homogeneity, it thus follows that for all $x\in P$, we have $T_0(x,x)\geq \beta_+\|x\|^2$.
- Similarly, there exists a $\beta_- >0$ such that for all $x\in N$, we have $T_0(x,x)\leq -\beta_-\|x\|^2$.
Let $\alpha = \min(\beta_+,\beta_-)$. Now, consider any symmetric tensor $T\in\mathcal{S}(E)$ such that $\|T-T_0\|\leq \frac{1}{2}$ (the precise choice of norm is not important, but it is convenient to use a generalization of the operator norm, to multilinear maps): in particular it follows that for all $x\in E$, we have $|T(x,x)- T_0(x,x)|\leq \frac{\alpha}{2}\|x\|^2$. Then, we have for all $x\in P$,
\begin{align}
T(x,x)\geq T_0(x,x)- |T(x,x)-T_0(x,x)|\geq \alpha\|x\|^2- \frac{\alpha}{2}\|x\|^2=\frac{\alpha}{2}\|x\|^2,
\end{align}
while for all $x\in N$,
\begin{align}
T(x,x)\leq T_0(x,x)+|T(x,x)-T_0(x,x)|\leq -\alpha\|x\|^2+ \frac{\alpha}{2}\|x\|^2.
\end{align}
This shows $T$ is positive-definite on $P$ and negative definite on $N$, and hence has the same signature as $T_0$. We have thus shown that for every $T_0\in\mathcal{S}_{p,q}(E)$, there is an $\alpha>0$ such that the open ball around $T_0$ of radius $\alpha$ in $\mathcal{S}(E)$ is contained in $\mathcal{S}_{p,q}(E)$. This precisely means $\mathcal{S}_{p,q}(E)$ is open in $\mathcal{S}(E)$.
Now, here's the statement and proof for non-changing of the signature of tensor fields.
Theorem.
Let $M$ be a $C^1$ connected $n$-dimensional manifold and $g$ a continuous symmetric $(0,2)$ tensor-field which is pointwise non-degenerate. Then, $g$ has the same signature at each point.
Fix non-negative integers $(p,q)$ such that $p+q=n$ and let $\Omega_{p,q}$ be the set of points in $M$ where $g$ has signature $(p,q)$. I claim this is open (possibly empty): consider any point $a\in \Omega_{p,q}$, and fix a coordinate neighbourhood $U$ around $a$. Then, $g$ the coordinate representative of $g$ gives us a continuous map $U\to \mathcal{S}(\Bbb{R}^n)$ (symmetric $(0,2)$ tensors on $\Bbb{R}^n$). Then $\mathcal{S}_{p,q}(\Bbb{R}^n)$ (which is open by the lemma above) has an open preimage $\widetilde{U}$ in $U$ (due to continuity). Thus, we have shown that $\widetilde{U}\subset \Omega_{p,q}$, thereby showing its openness.
In fact the above argument being valid for all $(p,q)$ also shows that each $\Omega_{p,q}$ is closed in $M$: because pointwise non-degeneracy of the tensor field $g$ means that $M=\bigcup\limits_{p+q=n}\Omega_{p,q}$ is a disjoint union of open sets, so the complement of $\Omega_{p,q}$ is $\bigcup\limits_{\substack{p'+q'=n\\ (p',q')\neq (p,q)}}\Omega_{p',q'}$, which is a union of open sets.
Since each $\Omega_{p,q}$ is clopen, by connectedness of $M$ it is either empty or equals $M$. In other words, if we fix a single point $a\in M$ and let $(p_0,q_0)$ be the signature of $g(a)$, then $M= \Omega_{p_0,q_0}$, and all other $\Omega_{p,q}$'s are empty, i.e $g$ has signature $(p_0,q_0)$ at each point of $M$.
I have to admit, in light of this, your questions 1-3 don't really make sense to me.
Let's look at question 1 first. We can always focus attention to one connected component at a time (and I guess physically, we should only look at connected spacetimes, since otherwise we can't make any measurements, since we can't send curves/signals to the other components). So, now that we're on a connected manifold, the result above tells us that if we require atleast a continuity for a symmetric pointwise non-degenerate tensor field $g$, then its signature at one point fixes it for all other points. So, there is no more signature-changing that can occur.
In view of this, we might get the idea to investigate what happens if we allow for a sometimes degenerate tensor field. But then the Einstein equations themselves loose meaning. First of all, the Levi-Civita connection is only defined for non-degenerate tensor fields, meaning we can't talk about curvature. Ok, so now we're going to have to come up with a somewhat arbitrary connection $\nabla$ even if it is compatible with the tensor field $g$ (it probably won't be unique since $g$ is not non-degenerate).
- From the connection, we can define the curvature endomorphism $\mathrm{Riem}$, i.e the full $(1,3)$ tensor field.
- From here, we can contract to get the Ricci tensor field $R_{ab}= R^{i}_{\, aib}$ (mind you it need not be symmetric anymore, even if the connection $\nabla$ is torsion-free).
- we certainly can't make sense of the Ricci scalar curvature because we can't contract the indices of $R_{ab}$.
- Hence, the Einstein tensor field $G_{ab}= R_{ab}-\frac{1}{2}g_{ab}R$ makes no sense, so we can't make sense of the Einstein field equations $G_{ab}=8\pi T_{ab}$.
It seems that all we can do is consider the vacuum case where the usual Einstein equations (for non-degnerate metric) are equivalent (in spacetime dimension $n\geq 3$) to $R_{ab}=0$.
Hopefully this discussion also makes clear that the answer to (3) is: no, the answer doesn't change (unless you only want to consider vacuum because otherwise the equations themselves don't make sense).
Lastly, regarding Q2, the Cauchy data is not a Lorentzian metric, but rather a Riemannian metric, and an extra symmetric $(0,2)$ tensor field $k$ (which is the to-be second fundamental form), plus some constraint equations. Assuming the evolution preserves continuity (many of the standard well-posedness theorems make more assumptions at the level of (weighted) Sobolev spaces) the signature cannot change in evolution for the trivial reason that we're solving Einstein's equations, i.e we're solving for a non-degenerate metric, so the signature won't change.
Also, these well-posedness theorems arise due to Lorentzian signature, because we're essentially solving quasilinear wave equations (once we go to a particular gauge), and the well-posedness of waves very strongly uses "energy estimates" which pretty much mean you're bounding "energy" fluxes at a later time by ones at the initial time. So the fact that we have only one time direction and rest are space is pretty instrumental in this regard (note btw that for fully Riemannian signature, i.e for elliptic problems, the initial-value problem is not well-posed, that's why in electrostatics, we have a boundary-value problem instead... and if I remember correctly, I'm pretty sure that the initial value problem in other signatures is ill-posed in many settings).
