(1) Let $a, b, c$ be real numbers, and satisfy $a+b+c=15, a^{2}+b^{2}+c^{2}=100$, then the product of the maximum and minimum values of $a$ is $\qquad$ .
(1) $\frac{25}{3}$ Hint: Since
$$
100-a^{2}=b^{2}+c^{2} \geqslant \frac{1}{2}(b+c)^{2}=\frac{1}{2}(15-a)^{2},
$$

therefore, we have $3 a^{2}-30 a+25 \leqslant 0$. Since the maximum and minimum values of $a$ are the two roots of the equation $3 a^{2}-30 a+25=0$, and the maximum and minimum values of $a$ can be achieved, thus, by Vieta's formulas, the product of the maximum and minimum values of $a$ is $\frac{25}{3}$.