9. (This sub-question is worth 16 points)

Given the function $f(x)=a \sin x-\frac{1}{2} \cos 2 x+a-\frac{3}{a}+\frac{1}{2}, a \in \mathbf{R}$ and $a \neq 0$.
(1) If for any $x \in \mathbf{R}$, $f(x) \leqslant 0$, find the range of values for $a$;
(2) If $a \geqslant 2$, and there exists $x \in \mathbf{R}$ such that $f(x) \leqslant 0$, find the range of values for $a$.
Solution: (1) $f(x)=\sin ^{2} x+a \sin x+a-\frac{3}{a}$. The necessary and sufficient condition for $f(x) \leqslant 0$ to hold for any $x \in \mathbf{R}$ is
$$
\left\{\begin{array}{l}
g(-1)=1-\frac{3}{a} \leqslant 0, \\
g(1)=1+2 a-\frac{3}{a} \leqslant 0 .
\end{array}\right.
$$

Solving this, we get the range of $a$ as $(0,1]$ 8 points
(2) Since $a \geqslant 2$, we have $-\frac{a}{2} \leqslant-1$. Therefore, $f(x)_{\text {min }}=1-\frac{3}{a}$.

Thus, the necessary and sufficient condition for there to exist $x \in \mathbf{R}$ such that $f(x) \leqslant 0$ is
$1-\frac{3}{a} \leqslant 0$, solving this gives $0<a \leqslant 3$.