$\begin{array}{l}\quad \text { 3. } \arcsin \frac{1}{\sqrt{10}}+\arccos \frac{7}{\sqrt{50}}+\operatorname{arctg} \frac{7}{31} \\ +\operatorname{arcctg} 10=\end{array}$
3. $\frac{\pi}{4}$.

Let $z_{1}=3+i, z_{2}=7+i, z_{3}=31+7 i, z_{4}=10+i$.
Then $z_{1} z_{2} z_{3} z_{4}=5050(1+i)$.
Also, $\arcsin -\frac{1}{\sqrt{10}}, \arccos \frac{7}{\sqrt{50}}, \operatorname{arctg} \frac{7}{31}$,
$\operatorname{arcctg} 10$ all belong to $\left(0, \frac{\pi}{2}\right)$,
$$
\begin{array}{l}
\therefore \arg z_{1}=\arcsin \frac{1}{\sqrt{10}}, \arg z_{2}=\arccos \frac{7}{\sqrt{50}}, \\
\arg z_{3}=\operatorname{arctg} \frac{7}{31}, \arg z_{4}=\operatorname{arcctg} 10 . \\
\arg z_{1}+\arg z_{2}+\arg z_{3}+\arg z_{4} \\
=\arg \left(z_{1} z_{1} z_{3} z_{4}\right)=\frac{\pi}{4} .
\end{array}
$$