(1) In a non-obtuse $\triangle A B C$, $A B>A C, \angle B=45^{\circ}, O$ and $I$ are the circumcenter and incenter of $\triangle A B C$, respectively, and $\sqrt{2} O I=A B-A C$. Find $\sin A$.
From the given conditions and Euler's formula, we have
$$
\left(\frac{c-b}{\sqrt{2}}\right)^{2}=O I^{2}=R^{2}-2 R r
$$

From the well-known geometric relationship, we get
$$
\begin{aligned}
r= & \frac{c+a-b}{2} \tan \frac{B}{2}=\frac{c+a-b}{2} \tan \frac{\pi}{8}= \\
& \frac{\sqrt{2}-1}{2}(c+a-b)
\end{aligned}
$$

From (1) and (2) and the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$, we get
$$
1-2(\sin C-\sin B)^{2}=2(\sin A+\sin C-\sin B)(\sqrt{2}-1)
$$

Since
$$
\begin{array}{c}
\angle B=\frac{\pi}{4}, \sin B=\frac{\sqrt{2}}{2} \\
\sin C=\sin \left(\frac{3 \pi}{4}-\angle A\right)=\frac{\sqrt{2}}{2}(\sin A+\cos A)
\end{array}
$$

Therefore,
$$
\begin{array}{c}
2 \sin A \cos A-(2-\sqrt{2}) \sin A \\
\sqrt{2} \cos A+\sqrt{2}-1=0 \\
(\sqrt{2} \sin A-1)(\sqrt{2} \cos A-\sqrt{2}+1)=0
\end{array}
$$

Thus, $\sin A=\frac{\sqrt{2}}{2}$ or $\cos A=1-\frac{\sqrt{2}}{2}$, in which case
$$
\begin{array}{l} 
\sin A= \sqrt{1-\cos ^{2} A}=\sqrt{1-\left(1-\frac{\sqrt{2}}{2}\right)^{2}}= \\
\sqrt{\sqrt{2}-\frac{1}{2}}
\end{array}
$$

In summary, $\sin A=\frac{\sqrt{2}}{2}$ or $\sin A=\sqrt{\sqrt{2}-\frac{1}{2}}$. (Strictly speaking, these two values should be verified to satisfy the conditions)