3. 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 $\sin A$.
(1998 CMO Problem)
3. Let $AB=c, CA=b, BC=a$, and let the incircle with radius $r$ of $\triangle ABC$ touch the sides $BC, CA, AB$ at points $D, E, F$, respectively, then $\frac{ID}{BD}=\frac{r}{\frac{a+c-b}{2}}=\tan \frac{\pi}{8}=\sqrt{2}-1$, so $r=\frac{\sqrt{2}-1}{2}(a+c-b)$.

From $\sqrt{2} OH=AB-AC$ and Euler's theorem $OI^{2}=R^{2}-2Rr$ ($R$ being the circumradius of $\triangle ABC$), we have $2(R^{2}-2Rr)=(c-b)^{2}$, i.e., $2R^{2}-2(\sqrt{2}-1)(a+c-b)R=(c-b)^{2}$. Substituting $a=2R \cdot \sin A, b=2R \cdot \sin B, c=2R \cdot \sin C$ into the above equation, simplifying yields $2-2(\sqrt{2}-1) \cdot (2 \sin A + 2 \sin C - 2 \sin B) = (2 \sin C - 2 \sin B)^{2}$, i.e., $1-2(\sqrt{2}-1) \cdot \left[\sin A + \left(\frac{\sqrt{2}}{2} \sin A + \frac{\sqrt{2}}{2} \cos A\right) - \frac{\sqrt{2}}{2}\right] = 2\left[\left(\frac{\sqrt{2}}{2} \sin A + \frac{\sqrt{2}}{2} \cos A\right) - \frac{\sqrt{2}}{2}\right]^{2}$, which simplifies to $2 \sin A \cdot \cos A + (\sqrt{2}-2) \cdot \sin A - \sqrt{2} \cos A + (\sqrt{2}-1) = 0$, i.e., $(\sqrt{2} \sin A - 1) \cdot [\sqrt{2} \cos A - (\sqrt{2}-1)] = 0$, thus $\sin A = \frac{\sqrt{2}}{2}$ or $\cos A = \frac{1}{2}(2-\sqrt{2})$. From $\cos A = \frac{1}{2}(2-\sqrt{2})$, we find $\sin A = \frac{1}{2} \sqrt{4 \sqrt{2}-2}$, hence $\sin A = \frac{\sqrt{2}}{2}$ or $\frac{1}{2} \sqrt{4 \sqrt{2}-2}$.