Four. (20 points) 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 $\sin \beta=m \cos (\alpha+\beta) \cdot \sin \alpha(m>0$, $\alpha+\beta \neq \frac{\pi}{2}$ ), 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 $\tan (\alpha+\beta)=(m+1) \tan \alpha$.
And $\tan \beta=\tan [(\alpha+\beta)-\alpha]$
$$
=\frac{m \tan \alpha}{1+(m+1) \tan ^{2} \alpha}.
$$

Therefore, $y=\frac{m x}{1+(m+1) x^{2}}$.
(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 $1 \leqslant x_{1}0$, so,
$$
u\left(x_{1}\right)<u\left(x_{2}\right).
$$

This shows that $u(x)=\frac{1}{m x}+\frac{m+1}{m} x$ is monotonically increasing on $[1,+\infty)$.
Hence, $f(x)_{\max }=\frac{m}{m+2}$.