The values of ${\tan }^2\frac{\alpha }{2}$, if $\sin \alpha +\cos \alpha =\frac{1}{5}$ is 

${\tan }^2\frac{\alpha }{2};\sin \alpha +\cos \alpha =\frac{1}{5}$

$\Rightarrow \frac{2\tan \frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}+\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}=\frac{1}{5}$

$\Rightarrow 5\left(1-{\tan }^2\frac{\alpha }{2}+2\tan \frac{\alpha }{2}\right)=1+{\tan }^2\frac{\alpha }{2}$

$\Rightarrow 5-5{\tan }^2\frac{\alpha }{2}+10\tan \frac{\alpha }{2}=1+{\tan }^2\frac{\alpha }{2}$

$\Rightarrow 6{\tan }^2\frac{\alpha }{2}-10\tan \frac{\alpha }{2}-4=0$

$\therefore \tan \frac{\alpha }{2}=\frac{10\pm \sqrt{100+16\times 6}}{12}$

$=\frac{10\pm \sqrt{196}}{12}$

$=\frac{10\pm 14}{12}$

$=2,-\frac{1}{3}$

$\therefore {\tan }^2\frac{\alpha }{2}=4,\frac{1}{9}$