9. (16 points) In the acute triangle $\triangle ABC$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, and $a, b, c$ form an arithmetic sequence, $\sin (A-C)=\frac{\sqrt{3}}{2}$. Find $\sin (A+C)$.
9. From $a 、 b 、 c$ forming an arithmetic sequence, we get
$$
a+c=2 b \text {. }
$$

By the Law of Sines, we know
$$
\sin A+\sin C=2 \sin B \text {. }
$$

In the acute $\triangle A B C$, we have
$$
\begin{array}{l}
\sin (A-C)=\frac{\sqrt{3}}{2} \\
\Rightarrow \angle A-\angle C=\frac{\pi}{3} .
\end{array}
$$

Since $\angle A+\angle B+\angle C=\pi$, then $\angle A=\frac{2 \pi}{3}-\frac{\angle B}{2}, \angle C=\frac{\pi}{3}-\frac{\angle B}{2}$
$\Rightarrow \sin \left(\frac{2 \pi}{3}-\frac{B}{2}\right)+\sin \left(\frac{\pi}{3}-\frac{B}{2}\right)=2 \sin B$
$\Rightarrow \sqrt{3} \cos \frac{B}{2}=2 \sin B$
$\Rightarrow \sin \frac{B}{2}=\frac{\sqrt{3}}{4}$.
Thus, $\cos \frac{B}{2}=\frac{\sqrt{13}}{4} \Rightarrow \sin B=\frac{\sqrt{39}}{8}$.
Therefore, $\sin (A+C)=\sin B=\frac{\sqrt{39}}{8}$.