13 Given $\tan (\alpha+\beta)=-2, \tan (\alpha-\beta)=\frac{1}{2}$, find $\frac{\sin 2 \alpha}{\sin 2 \beta}=$ $\qquad$ .
\begin{array}{l}13 \quad \frac{3}{5} . \quad \frac{\sin 2 \alpha}{\sin 2 \beta}=\frac{\sin [(\alpha+\beta)+(\alpha-\beta)]}{\sin [(\alpha+\beta)-(\alpha-\beta)]} \\ =\frac{\sin (\alpha+\beta) \cos (\alpha-\beta)+\cos (\alpha+\beta) \sin (\alpha-\beta)}{\sin (\alpha+\beta) \cos (\alpha-\beta)-\cos (\alpha+\beta) \sin (\alpha-\beta)} \\ =\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{\tan (\alpha+\beta)-\tan (\alpha-\beta)}=\frac{3}{5} . \\\end{array}