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 $\sqrt{2} \theta I=A B-A C$. Find $\sin A$.
Solution 1: From the known conditions and Euler's formula, we have
$$
\left(\frac{c-b}{\sqrt{2}}\right)^{2}=O r^{2}-R^{2}-2 R
$$

From the known geometric relationship, we get
$$
r=\frac{c+a-b}{2} \operatorname{tg} \frac{b}{2}=\frac{c+a-b}{2} \operatorname{tg} \frac{\pi}{8}
$$
$$
=\frac{\sqrt{2}-1}{2}(c+a-b) .
$$

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
$$
1-2(\sin C-\sin B)^{2}
=2(\sin A+\sin C-\sin B)(\sqrt{2}-1).
$$
Assuming $\angle B=\frac{\pi}{4}, \sin H=\frac{\sqrt{2}}{2}$,
$$
\sin C=\sin \left(\frac{3 \pi}{4}-\angle A\right)=\frac{\sqrt{2}}{2}(\sin A+\cos A),
$$
so,
$$
2 \sin A \cos A-(2-\sqrt{2}) \sin A
-\sqrt{2} \cos A+\sqrt{2}-1=0,
$$
$$
(\sqrt{2} \sin A \cdots 1)(\sqrt{2} \cos A-\sqrt{2}+1)=0.
$$
Thus, $\sin A=\frac{\sqrt{2}}{2}$ or $\cos A=1-\frac{\sqrt{2}}{2}$ (in this case, $\sin A=\sqrt{1-\cos ^{2} A}-\sqrt{1-\left(1-\frac{\sqrt{2}}{2}\right)^{2}}$
$$
=\sqrt{1 \sqrt{2}-\frac{1}{2}}
$$
Solving, we get $\sin A=\frac{V}{2}$ or $\sin A=\sqrt{\sqrt{2}-\frac{1}{2}}$.

Solution 2: Draw $H_{C} \perp A B, I_{*} \perp A, H_{A} \perp$ $R C, O D \perp B C$, where $f_{C} I_{C} 、 I_{A} 、 I_{A} 、 D$ are all feet, and $D$ is on the segment $I_{B} B>$.

Since $I_{A} B=\frac{1}{2} B C+I_{A} D, I_{A} C=\frac{1}{2} B C-I_{A} D$, we have $\sqrt{2} O I=2 I_{A} D$, i.e., $O I=\sqrt{2} I_{A} D$. $\frac{\pi}{4}$, so, either $O I \perp A B$, or $O I / / A B$.
We will discuss these cases separately.
(1) When $O I \perp A B$, point $I_{C}$ is the midpoint of $A B$, so $A I_{c}=B I_{C}$, i.e., $\frac{1}{2}(A B+A C$ $-B C)=\frac{1}{2}(A B+B C-$ $A C)$. Thus, $A C=B C$. Therefore, $\angle A=\angle B=\frac{\pi}{4}, \angle C=\frac{\pi}{2}$, and point $O$ coincides with $I_{C}$, as shown in Figure 1.
(2) When $O I / / A B$, the distance from point $O$ to $A B$ is exactly the radius $r$ of the incircle of $\triangle A B C$, i.e., $R \cos C=r=$ : $4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. So,
$$
\begin{array}{l}
\text { Therefore, } \\
\cos C=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\
=2 \sin \frac{\pi}{8}\left(2 \sin \frac{A}{2} \sin \frac{C}{2}\right) \\
=2 \sin \frac{\pi}{8}\left(-\cos \frac{A+C}{2}+\cos \frac{A-C}{2}\right) \\
=2 \sin \frac{\pi}{8}\left(-\sin \frac{\pi}{8}+\cos \left(\frac{3 \pi}{8}-\angle C\right)\right) \\
=\cos \frac{\pi}{4}-1+2 \sin \frac{\pi}{8} \cos \left(\frac{3 \pi}{8}-\angle C\right) \\
=\cos \frac{\pi}{4}-1+\sin \left(\frac{\pi}{2}-\angle C\right)+\sin \left(\angle C-\frac{\pi}{4}\right) .
\end{array}
$$

Eliminating $\cos C$ from both sides, we get $\sin \left(\angle C-\frac{\pi}{4}\right)=1-\frac{\sqrt{2}}{2}$.
Since $0<\angle C-\frac{\pi}{4}<\frac{\pi}{2}$ (because $\angle B=\frac{\pi}{4}$), we have
$$
\angle C-\frac{\pi}{4}=\arcsin \left(1-\frac{\sqrt{2}}{2}\right),
$$
$$
\angle C=\frac{\pi}{4}+\arcsin \left(1-\frac{\sqrt{2}}{2}\right),
$$
$$
\angle A=\frac{\pi}{2}-\arcsin \left(1-\frac{\sqrt{2}}{2}\right).
$$
Thus, $\sin A=\cos \left(\arcsin \left(1-\frac{\sqrt{2}}{2}\right)\right)$ $=\sqrt{1-\left(1-\frac{\sqrt{2}}{2}\right)^{2}}=\sqrt{\sqrt{2}-\frac{1}{2}}$, as shown in Figure 2.
In summary, either $\sin A=\frac{\sqrt{2}}{2}$, or $\sin A=\sqrt{\sqrt{2}-\frac{1}{2}}$.