9. (16 points) In $\triangle A B C$, $\overrightarrow{A B} \cdot \overrightarrow{A C}=8$, let $\angle B A C=\theta$, the area of $\triangle A B C$ is $S$, and it satisfies
$$
4(2-\sqrt{3}) \leqslant S \leqslant 4 \sqrt{3} \text { : }
$$
(1) Find the range of $\theta$;
(2) Find the maximum and minimum values of the function
$$
f(\theta)=2 \sqrt{3} \sin ^{2}\left(\frac{\pi}{4}+\theta\right)+2 \cos ^{2} \theta-\sqrt{3}
$$
(1) From $\overrightarrow{A B} \cdot \overrightarrow{A C}=8$, we get $|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \cos \theta=8$.
Since $4(2-\sqrt{3}) \leqslant \frac{1}{2}|\overrightarrow{A B}| \cdot|\overrightarrow{A C}| \sin \theta \leqslant 4 \sqrt{3}$,
it follows that $2-\sqrt{3} \leqslant \tan \theta \leqslant \sqrt{3}$.
Noting that $0<\theta<\pi$.
Therefore, the range of $\theta$ is $\left[\frac{\pi}{12}, \frac{\pi}{3}\right]$.
(2) Noting that
$$
\begin{array}{l}
f(\theta)=\sqrt{3}\left[1-\cos \left(\frac{\pi}{2}+2 \theta\right)\right]+(1+\cos 2 \theta)-\sqrt{3} \\
=\sqrt{3} \sin 2 \theta+\cos 2 \theta+1 \\
=2 \sin \left(2 \theta+\frac{\pi}{6}\right)+1
\end{array}
$$

Since $\frac{\pi}{12} \leqslant \theta \leqslant \frac{\pi}{3}$, it follows that $\frac{\pi}{3} \leqslant 2 \theta+\frac{\pi}{6} \leqslant \frac{5 \pi}{6}$.
Thus, when $2 \theta+\frac{\pi}{6}=\frac{\pi}{2}$, i.e., $\theta=\frac{\pi}{6}$,
$$
f(\theta)_{\max }=3 \text {; }
$$

When $2 \theta+\frac{\pi}{6}=\frac{5 \pi}{6}$, i.e., $\theta=\frac{\pi}{3}$, $f(\theta)_{\text {min }}=2$.