[Solution] (a) Let $\sin \alpha = x$, the $\alpha$ satisfying this equation must have the form $k \pi + (-1)^k \arcsin x$,

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 at most 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:
$$\sin \frac{\pi}{6}, \sin \frac{\pi}{3}, \sin \frac{7 \pi}{6}, \sin \frac{4 \pi}{3}.$$
(b) 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:
$$\begin{array}{l} \frac{1}{3}(\pi - \arcsin x) = \pi - \frac{1}{3}(2 \pi + \arcsin x) \\ \frac{1}{3}(3 \pi - \arcsin x) = \pi - \frac{1}{3} \arcsin x \\ \frac{1}{3}(5 \pi - \arcsin x) = 3 \pi - \frac{1}{3}(4 \pi + \arcsin x) \end{array}$$

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