1. Let $\alpha, \beta \in$ $\left(0, \frac{\pi}{2}\right)$, and $\sin ^{2} \alpha+\sin ^{2} \beta-\frac{\sqrt{6}}{2} \sin \alpha-\frac{\sqrt{10}}{2} \sin \beta+1=0$, then $\alpha+\beta=$ $\qquad$ $ـ$.
$$
\left(\sin \alpha-\frac{\sqrt{6}}{4}\right)^{2}+\left(\sin \beta-\frac{\sqrt{10}}{4}\right)^{2}=0 \Rightarrow \sin \alpha=\frac{\sqrt{6}}{4}, \sin \beta=\frac{\sqrt{10}}{4} \text {, }
$$

So $\sin \beta=\cos \alpha \Rightarrow \alpha+\beta=\frac{\pi}{2}$.