8. (58th Moscow Mathematical Olympiad, 1996) Given the value of $\sin \alpha$. Determine: (1) how many different values $\sin \frac{\alpha}{2}$ can have, (2) how many different values $\sin \frac{\alpha}{3}$ can have.
8. (1) Let $\sin \alpha=x$, then $\alpha$ must have the form
$$
k \pi+(-1)^{k} \arcsin x \text {, }
$$

where $k$ is an integer. Thus, the corresponding values of $\frac{\alpha}{2}$ correspond to 4 points on the unit circle:
$$
\begin{array}{l}
\frac{1}{2} \arcsin x, \frac{1}{2} \arcsin x+\pi . \\
\frac{1}{2}(\pi-\arcsin x), \frac{1}{2}(3 \pi-\arcsin x) .
\end{array}
$$

Therefore, $\sin \frac{\alpha}{2}$ can take 4 different values.
On the other hand, when $x=\frac{\sqrt{3}}{2}$, $\alpha=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{7 \pi}{3}, \frac{8 \pi}{3}, \cdots \sin \frac{\alpha}{2}$ can take 4 different values:
$$
\sin \frac{\pi}{6}, \sin \frac{\pi}{3}, \sin \frac{7 \pi}{6}, \sin \frac{4 \pi}{3} \text {. }
$$
(2) The corresponding values of $\frac{\alpha}{3}$ correspond to 6 points on the unit circle:
$$
\begin{array}{l}
\frac{1}{3} \arcsin x, \frac{1}{3}(2 \pi+\arcsin x), \frac{1}{3}(4 \pi+\arcsin x), \\
\frac{1}{3}(\pi-\arcsin x), \frac{1}{3}(3 \pi-\arcsin x), \frac{1}{3}(5 \pi-\arcsin x) .
\end{array}
$$

However, the last three points have the following relationships with the first three points:
$$
\begin{array}{l}
\frac{1}{3}(\pi-\arcsin x)=\pi-\frac{1}{3}(2 \pi+\arcsin x), \\
\left.\frac{1}{3}(3 \pi-\arcsin x)=\pi-\frac{1}{3} \arcsin x\right), \\
\frac{1}{3}(5 \pi-\arcsin x)=3 \pi-\frac{1}{3}(4 \pi+\arcsin x),
\end{array}
$$

so $\sin \frac{\alpha}{3}$ can take at most 3 different values.
Furthermore, when $\sin \alpha=0$, $\alpha=0,2 \pi, 4 \pi, 0, \cdots, \sin \frac{\alpha}{3}$ can take 3 different values:
$$
\sin 0, \sin \frac{2 \pi}{3}, \sin \frac{4 \pi}{3} \text {. }
$$