(1) From the problem, we have
$$
\begin{array}{l}
\cos (A-C)-\cos B=\frac{1}{2} \\
\Rightarrow \cos (A-C)+\cos (A+C)=\frac{1}{2} \\
\Rightarrow \cos A \cdot \cos C=\frac{1}{4} .
\end{array}
$$

Also, $b^{2}=a c$, then
$$
\sin ^{2} B=\sin A \cdot \sin C \text {. }
$$

Thus, $\frac{1}{4}-\sin ^{2} B$
$$
\begin{array}{l}
=\cos A \cdot \cos C-\sin A \cdot \sin C \\
=\cos (A+C)=-\cos B \\
\Rightarrow \frac{1}{4}-\left(1-\cos ^{2} B\right)=-\cos B \\
\Rightarrow 4 \cos ^{2} B+4 \cos B-3=0
\end{array}
$$
$\Rightarrow \cos B=\frac{1}{2}$ or $\cos B=-\frac{3}{2}$ (discard).
Since $0<\angle B<\frac{\pi}{2}$, thus, $\angle B=\frac{\pi}{3}$.
(2) From the conclusion of (1), (1) + (2) gives
$$
\cos (A-C)=\frac{1}{4}+\sin ^{2} B=1 \text {. }
$$

Then $\angle A=\angle C$.
Therefore, $\triangle A B C$ is an equilateral triangle.
Let the side length of $\triangle A B C$ be $x$. Then $0<x<5$. Hence
$$
\begin{array}{l}
\overrightarrow{A C} \cdot \overrightarrow{C D}=|\overrightarrow{A C}||\overrightarrow{C D}| \cos 60^{\circ}=\frac{1}{2} x(5-x) \\
=-\frac{1}{2}\left(\left(x-\frac{5}{2}\right)^{2}-\frac{25}{4}\right) \in\left(0, \frac{25}{8}\right],
\end{array}
$$

When and only when $x=\frac{5}{2}$, the above equality holds.
Thus, the range of $\overrightarrow{A C} \cdot \overrightarrow{C D}$ is $\left(0, \frac{25}{8}\right]$.
11. (13 points) In $\triangle A B C$, the sides $a, b, c$ opposite to $\angle A, \angle B, \angle C$ form a geometric sequence,
$$
2 \cos (A-C)-2 \cos B=1 \text {, }
$$

Extend $B C$ to point $D$ such that $B D=5$. Find:
(1) The size of $\angle B$;
(2) The range of $\overrightarrow{A C} \cdot \overrightarrow{C D}$.

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