5. Given $1-\cos \alpha-\cos \beta+\sin \alpha \cos \beta=0,1+\cos \alpha-\sin \beta+\sin \alpha \cos \beta=$ $0, \sin \alpha=$ $\qquad$
5. $\frac{1-\sqrt{10}}{3}$.
$$
1-\cos \alpha-\cos \beta+\sin \alpha \cos \beta=0, \cos \beta=\frac{1-\cos \alpha}{1-\sin \alpha} \text{ substitute into } 1+\cos \alpha-\sin \beta+\sin \alpha \cos \beta=0,
$$

we get $\sin \beta=2 \cos \alpha+\frac{1-\cos \alpha}{1-\sin \alpha}$. And $\cos ^{2} \beta+\sin ^{2} \beta=1$, simplifying and organizing we get $3 \sin ^{2} \alpha-2 \sin \alpha-3=0$, solving for $\sin \alpha=\frac{1-\sqrt{10}}{3}, \sin \alpha=\frac{1+\sqrt{10}}{3}$ (discard), $\sin \alpha=\frac{1-\sqrt{10}}{3}$.