9. (16 points) Given that the three interior angles of $\triangle A B C$ satisfy
$$
\begin{array}{l}
\angle A+\angle C=2 \angle B, \\
\cos A+\cos C=-\frac{\sqrt{2} \cos A \cdot \cos C}{\cos B} .
\end{array}
$$

Find the value of $\cos \frac{A-C}{2}$.
Given that
$$
\angle B=60^{\circ}, \angle A+\angle C=120^{\circ} \text {. }
$$

Let $\frac{\angle A-\angle C}{2}=\alpha$. Then
$$
\angle A-\angle C=2 \alpha \text {. }
$$

Thus, $\angle A=60^{\circ}+\alpha, \angle C=60^{\circ}-\alpha$.
Therefore, $\cos A+\cos C$
$$
\begin{array}{l}
=\cos \left(60^{\circ}+\alpha\right)+\cos \left(60^{\circ}-\alpha\right) \\
=2 \cos 60^{\circ} \cdot \cos \alpha=\cos \alpha,
\end{array}
$$
$$
\begin{aligned}
& \frac{\sqrt{2} \cos A \cdot \cos C}{\cos B} \\
= & \frac{\sqrt{2} \cos \left(60^{\circ}+\alpha\right) \cdot \cos \left(60^{\circ}-\alpha\right)}{\cos 60^{\circ}} \\
= & \sqrt{2}\left(\cos 2 \alpha+\cos 120^{\circ}\right) \\
= & 2 \sqrt{2}\left(\cos ^{2} \alpha-\frac{3}{4}\right) .
\end{aligned}
$$

Thus, $\cos \alpha=-2 \sqrt{2}\left(\cos ^{2} \alpha-\frac{3}{4}\right)$
$$
\begin{array}{l}
\Rightarrow 4 \sqrt{2} \cos ^{2} \alpha+2 \cos \alpha-3 \sqrt{2}=0 \\
\Rightarrow(2 \sqrt{2} \cos \alpha+3)(2 \cos \alpha-\sqrt{2})=0 .
\end{array}
$$

If $2 \sqrt{2} \cos \alpha+3=0$
$$
\Rightarrow \cos \alpha=-\frac{3 \sqrt{2}}{4}<-1 \text {. }
$$

Therefore, $2 \cos \alpha-\sqrt{2}=0 \Rightarrow \cos \alpha=\frac{\sqrt{2}}{2}$.