(Disclaimer: this is a follow-up question to Equation for Hubble Value as a function of time)
You still have a few misconceptions:
First, a model with a cosmological constant does lead to accelerated expansion. Look at the second derivative of $a(t)$ in my post:
$$
\ddot{a}(t) = \frac{1}{2}H_0^2\left(-2\,\Omega_{R,0}\,a(t)^{-3} - \Omega_{M,0}\,a(t)^{-2} +2\,\Omega_{\Lambda,0}\,a(t)\right).
$$
You see that $\ddot{a}(t)>0$ if $a(t)$ is sufficiently large. In particular, with the values
$$
H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\
\Omega_{R,0} \approx 0,\qquad\Omega_{M,0} = 0.315,\qquad\Omega_{\Lambda,0} = 0.685,\qquad\Omega_{K,0} = 0,
$$
you can work out that $\ddot{a}(t) > 0$ for $a(t)> 0.6$, which corresponds with $t > 7.7$ billion years. That is, the expansion began to accelerate when the universe was 7.7 billion years old, and will continue to do so. See also this post for more details: Can space expand with unlimited speed?
Second, a model with quintessence does have a big bang. The most general equation for $t(a)$, which I hadn't posted in my answer to your other question, is
$$
\begin{align}
t(a) &= \frac{1}{H_0}\int_0^a
\frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^{1-3w}}}.
\end{align}
$$
This is a well-behaved function when $a\rightarrow 0$, because $\Omega_{R,0}>0$. This means that in the early universe, radiation was the dominant factor (the other terms in the denominator of the integrand go to zero for $a\rightarrow 0$), and we get $t(0)=0$ or conversely $a(0)=0$.
In your previous question, you used a simplified model with $\Omega_{R,0} = \Omega_{M,0} = \Omega_{K,0} =0$. Those models don't have a big bang, because then the integrand becomes infinite for $a\rightarrow 0$. But of course, those toy models do not correspond with our actual universe. You need the general model.
So, both a model with a cosmological constant and one with quintessence produce a universe with a big bang and an accelerated expansion. Does the data suggest a non-constant dark energy? It's too soon to tell, the error-bars are still too large; a cosmological constant is still consistent with the data. It's obvious that quintessence provides a better fit, because it has the extra parameter $w$, but that itself doesn't mean that this extra parameter is necesary to explain the observations. But time will tell...
