9. (16 points) Given
$$
f(x)=\sin 2x + 3 \sin x + 3 \cos x \quad (0 \leqslant x < 2 \pi) \text{.}
$$
(1) Find the range of $f(x)$;
(2) Find the intervals of monotonicity of $f(x)$.
(1) Let
$\sin x+\cos x=t(-\sqrt{2} \leqslant t \leqslant \sqrt{2})$.
Then $\sin 2 x=t^{2}-1$.
Thus, we only need to find the range of $g(t)=t^{2}+3 t-1$.
Also, $g(t)=\left(t+\frac{3}{2}\right)^{2}-\frac{13}{4}$, so when $t= \pm \sqrt{2}$,
$g(t)$ attains its extreme values, which means the range of $f(x)$ is
$$
[1-3 \sqrt{2}, 1+3 \sqrt{2}] \text {. }
$$
(2) Notice that
$$
\begin{array}{l}
f^{\prime}(x)=2 \cos 2 x+3(\cos x-\sin x) \\
=(\cos x-\sin x)(2 \cos x+2 \sin x+3) .
\end{array}
$$

Since $2 \cos x+2 \sin x+3>0$, the decreasing interval of $f(x)$ is $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$, and the increasing intervals are
$$
\left[0, \frac{\pi}{4}\right],\left[\frac{5 \pi}{4}, 2 \pi\right] .
$$