Can an Electric field and a Magnetic field be non perpendicular to each other$?$ If yes, can an electromagnetic wave pass through that region$?$
I think that yes, both can be non perpendicular to each other as I haven't heard about any compulsion that both have to be perpendicular. Moreover, I guess, Transverse electric and magnetic fields are some good examples.
Very simply, you can just bring a charged insulator and a magnet together and place them into arbitrary positions. The $E$- and $B$-fields don't interact in this case, so they exist independently of one another. Note that I am in the regime of electrostatics and magnetostatics.
More specifically, place a magnet (approximated as a dipole) at the origin $\vec{r}_{0} = 0$ oriented such that its dipole moment is $\vec{\mu} = \mu\hat{z}$.
The magnetic field is
$$ \vec{B}(\vec{r}) = \frac{\mu_{0}}{4\pi}\frac{3(\vec{\mu}\cdot\hat{r})\hat{r} - \vec{\mu}}{r^{3}}. $$
Place an spherical insulator that is uniformly charged (approximated as a point charge) at $\vec{r}_{0} = 1\hat{z}$ with charge $q$.
The electric field is
$$ \vec{E}(\vec{r}) = \frac{q}{4\pi\epsilon_{0} |\vec{r}-\vec{r}_{0}|^{2}} \frac{\vec{r}-\vec{r}_{0}}{|\vec{r}-\vec{r}_{0}|}. $$
At position $\vec{r} = (0, 0, z)$,
\begin{align*}
\vec{E} = \frac{q}{4\pi\epsilon_{0} (z-1)^{2}}\hat{z} \quad\text{ and }\quad \vec{B} = \frac{\mu_{0}}{4\pi}\frac{2\mu}{z^{3}}\hat{z}.
\end{align*}
This situation clearly gives $\vec{E}\cdot\vec{B}\ne 0$ in a region $R$ around point $P$ located by $\vec{r}_{P} = (0, 0, z)$.
For the second part, I think that electromagnetic wave cannot pass through that region because Maxwell's equations tell us $$\vec{B}\cdot\vec{E}=0$$
Is my reasoning correct or is it not$?$ Is there anything else which I should know in regard of this problem$?$
This is wrong, because classical electromagnetism obeys the superposition principle for fields. This means if you have solution 1 and solution 2 and combine them, the two fields add linearly. At the classical level, electromagnetic fields can pass through any other region of $E$- and $B$-fields just fine.
In your case, go back to the example of a charged insulator and a bar magnet.
Assume both are locked into their positions (note that an insulator also locks its charges into place around it as well so they will not move around). We have $\vec{E}_{\text{static}}$ and $\vec{B}_{\text{static}}$ such that $\vec{E}_{\text{static}}\cdot\vec{B}_{\text{static}} \ne 0$ at some point $P$.
Now send an electromagnetic planewave with fields
$$ \vec{E}_{\text{rad}} = \hat{x}\, E_{0}\cos(kz-\omega t) \qquad\text{ and }\qquad \vec{B}_{\text{rad}} = \hat{y}\, B_{0}\cos(kz-\omega t), $$
which indeed satisfy $\vec{E}_{\text{rad}}\cdot\vec{B}_{\text{rad}} = 0$.
When the radiation enters region $R$, by the superposition principle, the fields are $\vec{E}_{\text{static}} + \vec{E}_{\text{rad}}$ and $\vec{B}_{\text{static}} + \vec{B}_{\text{rad}}$. Note that the fact that the magnet and insulator are fixed in place and the charges on the insulator are fixed as well is important here; this assumption is physically plausible and needed for our argument (otherwise the "static" fields will change).
The result is
\begin{align*}
(\vec{E}_{\text{static}} + \vec{E}_{\text{rad}})\cdot (\vec{B}_{\text{static}} + \vec{B}_{\text{rad}}) &=
\vec{E}_{\text{static}}\cdot\vec{B}_{\text{static}} +
\underbrace{ \vec{E}_{\text{static}}\cdot\vec{B}_{\text{rad}} }_{\hat{z}\cdot\hat{y}=0} +
\underbrace{ \vec{E}_{\text{rad}}\cdot\vec{B}_{\text{static}} }_{\hat{x}\cdot\hat{z}=0} +
\underbrace{ \vec{E}_{\text{rad}}\cdot\vec{B}_{\text{rad}} }_{\hat{x}\cdot\hat{y}=0} \\
&= \vec{E}_{\text{static}}\cdot\vec{B}_{\text{static}} \ne 0
\end{align*}
If you wish, you can replace the planewave by a localized wave pulse going in the $+z$-direction (linearly polarized the same way) and you would obtain the exact same result. By continuity, the fact that this is not $0$ at point $P$ means it is not zero at around a neighborhood $R$ of $P$ (i.e. open region around $P$).
The specific condition $\vec{E}\cdot\vec{B} = 0$ cannot be derived from Maxwell's equations in general unless you add more assumptions such as the class of solutions you are considering.
