Question 97, Solution: As shown in the figure, according to the given conditions, $\angle \mathrm{AOB}=90^{\circ}$. Let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}$, then $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$, and $\vec{a} \cdot \vec{b}=0$. Therefore, we know:
$$
\begin{array}{l}
\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}=\overrightarrow{\mathrm{c}} \cdot (\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})+(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{c}}) \cdot (\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}) \\
=|\overrightarrow{\mathrm{c}}|^{2}-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=1-2 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}
\end{array}
$$

On one hand, $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}} \leq 1$, when $\mathrm{C}$ is infinitely close to point $\mathrm{A}$ on the major arc $\mathrm{ACB}$, the equality holds; on the other hand, $\vec{a} \cdot \vec{c} \geq -1$, hence $\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}} \leq 3$, when $C$ is the intersection point of ray $A O$ and the major arc $A B$, the equality holds.
In summary, the range of $\overrightarrow{\mathrm{OC}} \cdot \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}$ is $(-1,3)$.
Question 97, In $\triangle ABC$, $AB=\sqrt{2}, \angle ACB=\frac{\pi}{4}, O$ is the circumcenter of $\triangle ABC$, try to find the range of $\overrightarrow{OC} \cdot \overrightarrow{AB} + \overrightarrow{CA} \cdot \overrightarrow{CB}$.