For the problem: "Let the function $f(x)=a x^{2}+b x$, and $1 \leqslant f(-1) \leqslant 2, 2 \leqslant f(1) \leqslant 4$, find the range of $f(-2)$", a student provided the following solution. Please point out the errors in this solution and provide the correct solution.
Incorrect solution: From the given conditions,
$$
\begin{array}{c}
1 \leqslant a-b \leqslant 2, \\
2 \leqslant a+b \leqslant 4 \\
-2 \leqslant b-a \leqslant-1 .
\end{array}
$$

By adding (1) and (2), we get $\frac{3}{2} \leqslant a \leqslant 3$; by adding (2) and (3), we get $0 \leqslant b \leqslant \frac{3}{2}$.
Therefore, $3 \leqslant 4 a-2 b \leqslant 12$, which means $3 \leqslant f(-2) \leqslant 12$.
(4) The mistake lies in determining the range of $a$ and $b$; $a$ and $b$ cannot both take the maximum or minimum values simultaneously. The correct range for $f(-2)$ is $(3,12)$