11. In $\triangle A B C$, the sides opposite to the internal angles $A, B, C$ are $a, b, c$ respectively. Given that $\sin ^{2} A+\sin ^{2} B-\sin ^{2} C=$ $\sqrt{2} \sin A \sin B$.
(1) Find $C$;
(2) If $\cos B=\frac{3}{5}$, and $D$ is a point on side $B C$ such that $C D=4 B D$, and the area of $\triangle A C D$ is $\frac{7}{5}$, find $A C$.
(1) $a^{2}+b^{2}-c^{2}=\sqrt{2} a b \Rightarrow \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{\sqrt{2}}{2} \Rightarrow C=45^{\circ}$.
(2) By the Law of Sines, $\frac{b}{c}=\frac{\sin B}{\sin C}=\frac{4 \sqrt{2}}{5}$, and $\sin A=\left(\frac{4}{5}+\frac{3}{5}\right) \cdot \frac{\sqrt{2}}{2}=\frac{7 \sqrt{2}}{10}$,
$S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{7}{5} \cdot \frac{5}{4}=\frac{7}{4} \Rightarrow b c \sin A=\frac{7 \sqrt{2}}{10} b c=\frac{7}{2} \Rightarrow b c=\frac{5 \sqrt{2}}{2}$.
Therefore, $b^{2}=\frac{4 \sqrt{2}}{5} \cdot \frac{5 \sqrt{2}}{2}=4 \Rightarrow b=2$.