2. 2. $47{ }^{\star \star}$ Given acute angles $\alpha, \beta$ satisfy
$$
\sin \beta=m \cos (\alpha+\beta) \cdot \sin \alpha\left(m>0, \alpha+\beta \neq \frac{\pi}{2}\right) .
$$

If $x=\tan \alpha, y=\tan \beta$,
(1) Find the expression for $y=f(x)$;
(2) Under (1), when $\alpha \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$, find the maximum value of the function $y$.
(1) From $\sin \beta=m \cos (\alpha+\beta) \cdot \sin \alpha\left(m>0, \alpha+\beta \neq \frac{\pi}{2}\right)$, we have $\sin [(\alpha+$ $\beta)-\alpha]=m \cos (\alpha+\beta) \cdot \sin \alpha$, which means
$$
\sin (\alpha+\beta) \cdot \cos \alpha=(m+1) \cos (\alpha+\beta) \cdot \sin \alpha .
$$

Since $\alpha, \beta$ are acute angles, and $\alpha+\beta \neq \frac{\pi}{2}$, we have $\tan (\alpha+\beta)=(m+1) \tan \alpha$. Therefore, $\tan \beta=$ $\tan [(\alpha+\beta)-\alpha]=\frac{m \tan \alpha}{1+(m+1) \tan ^{2} \alpha}$, so $y=\frac{m x}{1+(m+1) x^{2}}$.
(2) From (1), we know
$$
y=\frac{m x}{1+(m+1) x^{2}}=\frac{1}{\frac{1}{m x}+\left(\frac{1+m}{m}\right) x}(x \geqslant 1) .
$$

Let $u(x)=\frac{1}{m x}+\frac{m+1}{m} x$, and suppose $1 \leqslant x_{1}<x_{2}$, then we have
$$
u\left(x_{1}\right)-u\left(x_{2}\right)=\frac{x_{1}-x_{2}}{m x_{1} x_{2}}\left[(m+1) x_{1} x_{2}-1\right]<0,
$$

which means $u\left(x_{1}\right)<u\left(x_{2}\right)$. This indicates that $u(x)=\frac{1}{m x}+\frac{m+1}{m} x$ is monotonically increasing on $[1,+\infty)$. Therefore,
$$
f(x)_{\max }=\frac{m}{m+2}
$$