9. If $x_{1}, x_{2}$ are the real roots of the equation $x^{2}+a x+a-\frac{1}{2}=0$.
(1) Find the set of real values $A$ for $a$;
(2) Is there a real number $m$ such that the inequality $m^{2}+t m+4 \sqrt{2}+6 \geqslant\left(x_{1}-3 x_{2}\right)\left(x_{2}-3 x_{1}\right)$ holds for any $a \in A$ and $t \in[-1,1]$? If it exists, find the range of $m$; if not, explain the reason.
9. Solution: (1) $\because x^{2}+a x+a-\frac{1}{2}=0$ has two real roots, $\therefore \Delta=a^{2}-4\left(a-\frac{1}{2}\right) \geqslant 0$, i.e., $a^{2}-4 a+2 \geqslant 0$.
$$
A=\{a \mid a \in(-\infty, 2-\sqrt{2}] \cup[2+\sqrt{2},+\infty)\} .
$$
(2) Since $x_{1}+x_{2}=-a, \quad x_{1} x_{2}=a-\frac{1}{2}$,
$$
\begin{aligned}
f=\left(x_{1}-3 x_{2}\right)\left(x_{2}-3 x_{1}\right)=16 x_{1} x_{2}-3\left(x_{1}+x_{2}\right)^{2}, \\
f(a)=16\left(a-\frac{1}{2}\right)-3(-a)^{2}=-3 a^{2}+16 a-8,
\end{aligned}
$$

Since the axis of symmetry is $a=\frac{8}{3}$, thus $f(a)_{\max }=f(2+\sqrt{2})=6+4 \sqrt{2}$.
Convert the problem into for any $t \in[-1,1]$, the inequality $m^{2}+t m+4 \sqrt{2}+6 \geqslant 6+4 \sqrt{2}$ always holds. That is, $m^{2}+t m \geqslant 0$. Let $g(t)=m^{2}+t m$, to make the inequality $m^{2}+t m \geqslant 0$ hold for any $t \in[-1,1]$.
It is equivalent to $\left\{\begin{array}{l}g(-1) \geqslant 0 \\ g(1) \geqslant 0\end{array} \Rightarrow\left\{\begin{array}{l}m^{2}-m \geqslant 0 \\ m^{2}+m \geqslant 0\end{array} \Rightarrow m \geqslant 1\right.\right.$ or $m \leqslant-1$ or $m=0$.