8. 5 (CMO 13) In a non-obtuse $\triangle ABC$, $AB > AC$, $\angle B = 45^{\circ}$, $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, and $\sqrt{2} OI = AB - AC$. Find $\sin A$.
8. 5 From the given conditions and Euler's formula $\left(\frac{c-b}{\sqrt{2}}\right)^{2}=O I^{2}=R^{2}-2 R r$. Using a well-known geometric relationship,
$$
r=\frac{c+a-b}{2} \tan \frac{B}{2}=\frac{c+a-b}{2} \tan \frac{\pi}{8}=\frac{\sqrt{2}-1}{2}(c+a-b) .
$$

Applying the Law of Sines, eliminating $r, a, b, c$ and $R$ from the above equation yields: $1-2(\sin C-\sin B)^{2}=2(\sin A+\sin C-\sin B)$ $(\sqrt{2}-1), \sin B=\sqrt{2} / 2$, substituting $C=3 \pi / 4-A$ into the equation, it simplifies to a trigonometric equation in terms of angle $A$.
Solving this, we get $\sin A=\frac{\sqrt{2}}{2}$, or $\sin A=\sqrt{\sqrt{2}-\frac{1}{2}}$.