已知 $\alpha$ 是一个三角形的内角, 且 $\sin \alpha+\cos \alpha=\frac{1}{5}$

(1) 求 $\tan \alpha$ 的值;

(2) 用 $\tan \alpha$ 表示 $\frac{1}{\sin ^{2} \alpha-\cos ^{2} \alpha}$ 并求其值.
) 将 $\sin \alpha+\cos \alpha=\frac{1}{5}$, 两边平方得: $(\sin \alpha+\cos \alpha)^{2}=1+2 \sin \alpha \cos \alpha=\frac{1}{25}$,

$\therefore \sin \alpha \cos \alpha=-\frac{12}{25}<0$, 即 $\sin \alpha>0, \cos \alpha<0$,

与 $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, 联立得: $\left\{\begin{array}{l}\sin \alpha+\cos \alpha=\frac{1}{5} \\ \sin ^{2} \alpha+\cos ^{2} \alpha=1\end{array}, \quad \therefore\left\{\begin{array}{l}\sin \alpha=\frac{4}{5} \\ \cos \alpha=-\frac{3}{5}\end{array}\right.\right.$,

则 $\tan \alpha=-\frac{4}{3}$;

(2) $\frac{1}{\sin ^{2} \alpha-\cos ^{2} \alpha}=\frac{\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin ^{2} \alpha-\cos ^{2} \alpha}=\frac{\tan ^{2} \alpha+!}{\tan ^{2} \alpha-1}=\frac{\left(-\frac{5}{3}\right)^{2}+1}{\left(-\frac{4}{3}\right)^{2}-1}=\frac{25}{7}$.