3. Given $\arcsin (\sin \alpha+\sin \beta)+\arcsin (\sin \alpha-\sin \beta)=\frac{\pi}{2}, \sin ^{2} \alpha+\sin ^{2} \beta=$
3. $\frac{1}{2}$.

From the given, $\arcsin (\sin \alpha+\sin \beta)=\frac{\pi}{2}-\arcsin (\sin \alpha-\sin \beta) \Rightarrow \sin \alpha+\sin \beta=\cos (\arcsin (\sin \alpha- \sin \beta))=\sqrt{1-(\sin \alpha-\sin \beta)^{2}} \Rightarrow(\sin \alpha+\sin \beta)^{2}=1-(\sin \alpha-\sin \beta)^{2} \Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta=\frac{1}{2}$