58. 13. Given the value of $\sin \alpha$. How many different values can $\left.a) \sin \frac{\alpha}{2}, b\right)$ $\sin \frac{\alpha}{3}$ have at most, respectively?
58. 13. a) Let $\sin \alpha=x$, the $\alpha$ that satisfies this must!! make $\frac{\alpha}{2}$ correspond to 4 points on the unit circle:
$$
\begin{array}{l}
\frac{1}{2} \arcsin x, \frac{1}{2} \arcsin x+\pi, \\
\frac{1}{2}(\pi-\arcsin x), \frac{1}{2}(3 \pi-\arcsin x) .
\end{array}
$$

Thus, $\sin \frac{\alpha}{2}$ can take at most 4 different values.
On the other hand, when $x=\frac{\sqrt{3}}{2}, \alpha=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{7 \pi}{3}, \frac{8 \pi}{3} \pi .5$, $\sin \frac{\alpha}{2}$ can take 4 different values. Therefore, the answer is 4.
b) The reasoning process is similar, $\frac{\alpha}{3}$ has the form $\frac{1}{3}(k \pi$
$$
\begin{array}{l}
\frac{1}{3} \arcsin x,-\frac{1}{3}(2 \pi+\arcsin x), \\
-\frac{1}{3}(4 \pi+\arcsin x), \frac{1}{3}(\pi-\arcsin x), \\
\frac{1}{3}(3 \pi-\arcsin x), \frac{1}{3}(5 \pi-\arcsin x) .
\end{array}
$$

By the induced formulas of the sine function, the sine values of the 1st, 2nd, 4th, 3rd, and 6th points are equal, so $\sin \frac{\alpha}{3}$ can take no more than 3 different values. It can achieve 3 different values. Therefore, the answer is 3.