We have, $\tan \alpha=\frac{1}{7}$
$$
\begin{array}{cc}
\Rightarrow & \alpha=\tan ^{-1} \frac{1}{7} \text { and } \sin \beta=\frac{1}{\sqrt{10}} \\
\Rightarrow & \tan \beta=\frac{1}{3} \quad\left[\because \tan \beta=\frac{\sin }{\sqrt{1-\sin ^2 \beta}}\right] \\
\Rightarrow & \beta=\tan ^{-1} \frac{1}{3}
\end{array}
$$

Now, $\alpha+2 \beta=\tan ^{-1} \frac{1}{7}+2 \tan ^{-1} \frac{1}{3}$
$$
\begin{gathered}
=\tan ^{-1} \frac{1}{7}+\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}\right) \\
{\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]} \\
=\tan ^{-1} \frac{1}{7}+\tan ^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right)=\tan ^{-1} \frac{1}{7}+\tan ^{-1}\left(\frac{6}{8}\right) \\
=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{3}{4}=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \cdot \frac{3}{4}}\right)
\end{gathered}
$$


$$
\begin{aligned}
& =\tan ^{-1}\left(\frac{\frac{4+21}{28}}{\frac{28-3}{28}}\right)=\tan ^{-1}\left(\frac{\frac{25}{28}}{\frac{28}{28}}\right)=\tan ^{-1}(1)=\frac{\pi}{4} \\
& =45^{\circ} \quad\left[\because \tan ^{-1}(1)=45^{\circ}\right]
\end{aligned}
$$