13. Let $x, y$ be non-zero real numbers, and satisfy $\frac{x \sin \frac{\pi}{5} + y \cos \frac{\pi}{5}}{x \cos \frac{\pi}{5} - y \sin \frac{\pi}{5}} = \tan \frac{9 \pi}{20}$.
(1) Find the value of $\frac{y}{x}$;
(2) In $\triangle ABC$, if $\tan C = \frac{y}{x}$, find the maximum value of $\sin 2A + 2 \cos B$.
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\begin{array}{l}
\text { (1) } \frac{x \tan \frac{\pi}{5}+y}{x-y \tan \frac{\pi}{5}}=\tan \frac{9 \pi}{20} \Rightarrow \frac{\tan \frac{\pi}{5}+\frac{y}{x}}{1-\frac{y}{x} \cdot \tan \frac{\pi}{5}}=\tan \left(\frac{\pi}{5}+\frac{\pi}{4}\right)=\frac{\tan \frac{\pi}{5}+1}{1-\tan \frac{\pi}{5}} \\
\Rightarrow\left(\tan \frac{\pi}{5}+\frac{y}{x}\right)\left(1-\tan \frac{\pi}{5}\right)=\left(\tan \frac{\pi}{5}+1\right)\left(1-\frac{y}{x} \cdot \tan \frac{\pi}{5}\right) \\
\Rightarrow \frac{y}{x}\left(\tan ^{2} \frac{\pi}{5}+1\right)=\tan ^{2} \frac{\pi}{5}+1 \Rightarrow \frac{y}{x}=1 .
\end{array}
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(2) \( C=45^{\circ} \Rightarrow \sin 2 A+2 \cos B=\sin 2 A+2 \cos \left(135^{\circ}-A\right)=\sin 2 A-\sqrt{2} \cos A+\sqrt{2} \sin A \), let \( t=\cos A-\sin A=\sqrt{2} \cos \left(A+45^{\circ}\right) \in(-\sqrt{2}, 1) \), then \( \sin 2 A=1-t^{2} \), so \( \sin 2 A+2 \cos B=1-t^{2}-\sqrt{2} t=-\left(t+\frac{\sqrt{2}}{2}\right)^{2}+\frac{3}{2} \leqslant \frac{3}{2} \), equality holds when \( A=75^{\circ}, B=60^{\circ} \). In summary, the maximum value of \( \sin 2 A+2 \cos B \) is \( \frac{3}{2} \).
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