4. Rewrite the original system of equations in the form

$$
\left\{\begin{array}{l}
a=-3 x^{2}+5 x-2 \\
(x+2) a=4\left(x^{2}-1\right)
\end{array}\right.
$$

Notice that $x=-2$ is not a solution to the second equation of the system. Therefore, the system can be rewritten as

$$
\left\{\begin{array}{l}
a=-3 x^{2}+5 x-2 \\
a=\frac{4\left(x^{2}-1\right)}{x+2}
\end{array}\right.
$$

If the system is consistent, then its solution satisfies the equation

$$
\frac{4\left(x^{2}-1\right)}{x+2}=-3 x^{2}+5 x-2
$$

Let's solve this equation. For this, we factorize the left and right sides of the equation:

$$
\frac{4(x-1)(x+1)}{x+2}=(x-1)(-3 x+2)
$$

The last equation can be reduced to the form

$$
x(x-1)(3 x+8)=0
$$
The solution $x=0$ corresponds to the parameter $a=-2$, the solution $x=1$ corresponds to the parameter $a=0$, and $x=-\frac{8}{3}$ corresponds to the parameter $a=-\frac{110}{3}$.

Answer: 1) The system is consistent for $a=-2,0,-\frac{110}{3},-24$;

2) for $a=0 \quad x=1$; for $a=-2 \quad x=0$; for $a=-\frac{110}{3}$

$$
x=-\frac{8}{3}
$$