9. (16 points) Let the constant $a \in \mathbf{R}$, and the function
$$
f(x)=(a-x)|x|
$$

has an inverse function $f^{-1}(x)$. If the inequality
$$
f^{-1}\left(x^{2}+m\right)<f(x)
$$

holds for all $x \in[-2,2]$, find the range of the real number $m$.
II. 9. Since $f(0)=f(a)=0$ and there exists an inverse function $f^{-1}(x)$, therefore, $a=0, f(x)=-x|x|$.
Obviously, the function $f(x)$ is decreasing on $\mathbf{R}$.
Thus, $f^{-1}\left(x^{2}+m\right)f(f(x))$,

i.e.,
$$
\begin{array}{l}
x^{2}+m>f(f(x)) \\
\Rightarrow m>-x^{2}+f(f(x))
\end{array}
$$

holds for all $x \in[-2,2]$.
$$
\begin{array}{l}
\text { Let } g(x)=-x^{2}+f(f(x)) \\
=-x^{2}+x^{3}|x| \quad (x \in[-2,2]) \text {. }
\end{array}
$$

Notice that,
$$
\begin{array}{l}
g(x)=-x^{2}+x^{3}|x| \leqslant-x^{2}+x^{4} \\
=\left(x^{2}-\frac{1}{2}\right)^{2}-\frac{1}{4} \leqslant 12 .
\end{array}
$$

The equality holds when $x=2$.
Thus, $g(x)_{\max }=12$.
Consequently, $m \in(12,+\infty)$.