(1)已知 $\sin \alpha+\cos \alpha=\frac{4}{5}, 0<\alpha<\pi$, 求 $\sin \alpha-\cos \alpha$ ；

(2)已知 $\tan \alpha=2$, 求 $\frac{2 \sin \alpha-\cos \alpha}{\sin \alpha+3 \cos \alpha}$.
(1) $: \because \sin \alpha+\cos \alpha={ }^{\frac{4}{5}}, \therefore(\sin \alpha+\cos \alpha)^{2}={ }^{\frac{16}{25}}$,

$\therefore 2 \sin \alpha \cos \alpha=-{ }^{\frac{9}{25}}, \therefore(\sin \alpha-\cos \alpha)^{2}=1-2 \sin \alpha \cos \alpha={ }^{\frac{34}{25}}$, 又 $: 0<\alpha<\pi$ 且 $2 \sin \alpha \cos \alpha=-{ }^{\frac{9}{25}}$,
$\therefore:^{\frac{\pi}{2}}<\alpha<\pi, \therefore \sin \alpha-\cos \alpha=\frac{\sqrt{34}}{5}$.

$(2)^{\frac{2 \sin \alpha-\cos \alpha}{\sin \alpha+3 \cos \alpha}=\frac{2 \tan \alpha-1}{\tan \alpha+3}}={ }^{2+3}={ }^{\frac{2 \times 2-1}{5}}=\frac{3}{5}$.