(a) Since $\sin \beta = \frac{1}{{\sqrt {10} }}$

$\Rightarrow \tan \beta = \frac{1}{3}$

==> $\tan 2\beta = \frac{{2\tan \beta }}{{1 – {{\tan }^2}\beta }} = \frac{3}{4}$

$\therefore \tan (\alpha + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 – \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$

Now, $0 < \beta < \frac{\pi }{2}$ and $\tan 2\beta = \frac{3}{4} > 0$ both 

==> $0 < 2\beta < \frac{\pi }{2}$. 

Again,$0 < \alpha < \frac{\pi }{2}$ and $0 < 2\beta < \frac{\pi }{2}$ both 

==> $0 < \alpha + 2\beta < \pi $

Thus, $0 < \alpha + 2\beta < \pi $ and $\tan (\alpha + 2\beta ) = 1$ both 

==> $\alpha + 2\beta = \frac{\pi }{4} $

$\Rightarrow 2\beta = \frac{\pi }{4} – \alpha $.