1. Given $\alpha, \beta$ are acute angles, $\cos \alpha=\frac{\sqrt{5}}{5}, \cos \beta=$ $\frac{\sqrt{10}}{10}$. Then $\alpha+\beta=$ $\qquad$ .
$$
\text { 2. } \frac{3 \pi}{4} \text {. }
$$

From the given, $\sin \alpha=\frac{2 \sqrt{5}}{5}, \sin \beta=\frac{3 \sqrt{10}}{10}$, we get
$$
\begin{array}{l}
\cos (\alpha+\beta)=\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta \\
=\frac{\sqrt{5}}{5} \times \frac{\sqrt{10}}{10}-\frac{2 \sqrt{5}}{5} \times \frac{3 \sqrt{10}}{10}=-\frac{\sqrt{2}}{2} .
\end{array}
$$

Therefore, $\alpha+\beta=\frac{3 \pi}{4}$.