One. (20 points) 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 $\sin 2A + 2 \cos B$.
(1) Let $\frac{y}{x}=\tan \theta$. Then from the given information,
$$
\begin{array}{c}
\frac{\tan \frac{\pi}{5}+\tan \theta}{1-\tan \theta \cdot \tan \frac{\pi}{5}}=\tan \frac{9 \pi}{20} \\
\Rightarrow \tan \left(\theta+\frac{\pi}{5}\right)=\tan \frac{9 \pi}{20} \\
\Rightarrow \theta+\frac{\pi}{5}=k \pi+\frac{9 \pi}{20} \\
\Rightarrow \theta=k \pi+\frac{\pi}{4}(k \in \mathbf{Z}) .
\end{array}
$$
$$
\text { Hence } \frac{y}{x}=\tan \theta=\tan \left(k \pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4}=1 \text {. }
$$
(2) From (1), we get $\tan C=1$.

Since $0<\angle C<\pi$, therefore, $\angle C=\frac{\pi}{4}$.
Then $\angle A+\angle B=\frac{3 \pi}{4} \Rightarrow 2 \angle A=\frac{3 \pi}{2}-2 \angle B$.
Thus, $\sin 2 A+2 \cos B=\sin \left(\frac{3 \pi}{2}-2 B\right)+2 \cos B$
$$
\begin{array}{l}
=-\cos 2 B+2 \cos B=-2 \cos ^{2} B+2 \cos B+1 \\
=-2\left(\cos B-\frac{1}{2}\right)^{2}+\frac{3}{2} .
\end{array}
$$

Therefore, when $\cos B=\frac{1}{2}$, i.e., $\angle B=\frac{\pi}{3}$, $\sin 2 A+$ $2 \cos B$ achieves its maximum value of $\frac{3}{2}$.