19.已知 $\frac{\sin \alpha+3 \cos \alpha}{3 \cos \alpha-\sin \alpha}=5$, 则 $\tan \alpha=$ , $\sin ^{2} \alpha-\sin \alpha \cos \alpha=$ \$ \qquad \$ .
由$\frac{\sin{\alpha}+3\cos{\alpha}}{3\cos{\alpha}-\sin{\alpha}}=5$,可得$\frac{\tan{\alpha}+3}{3-\tan{\alpha}}=5$,解得$\tan{\alpha}=2$,故$\sin^{2}{\alpha}-\sin{\alpha}\cos\alpha=\frac{\sin ^{2} \alpha-\sin \alpha \cos \alpha}{\sin ^{2} \alpha+\cos ^{2} \alpha}=\frac{\tan ^{2} \alpha-\tan \alpha}{\tan ^{2} \alpha+1}=\frac{2}{5}$.