1. Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, and
$$
\sin \beta=2 \cos (\alpha+\beta) \cdot \sin \alpha\left(\alpha+\beta \neq \frac{\pi}{2}\right) \text {. }
$$

then the maximum value of $\tan \beta$ is
$$
\begin{array}{l}
\text { II, 1. } \frac{\sqrt{3}}{3} . \\
\sin [(\alpha+\beta)-\alpha]=2 \cos (\alpha+\beta) \cdot \sin \alpha, \\
\sin (\alpha+\beta) \cdot \cos \alpha-\cos (\alpha+\beta) \cdot \sin \alpha \\
=2 \cos (\alpha+\beta) \cdot \sin \alpha, \\
\sin (\alpha+\beta) \cdot \cos \alpha=3 \cos (\alpha+\beta) \cdot \sin \alpha,
\end{array}
$$

Then $\tan (\alpha+\beta)=3 \tan \alpha$.
$$
\begin{array}{l}
\tan \beta=\tan [(\alpha+\beta)-\alpha]=\frac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \cdot \tan \alpha} \\
=\frac{2}{\tan \alpha}+3 \tan \alpha \leqslant \frac{2}{2 \sqrt{\frac{1}{\tan \alpha} \cdot 3 \tan \alpha}} .
\end{array}
$$

Therefore, $\tan \beta \leqslant \frac{\sqrt{3}}{3}$.