5. After substituting $x=\cos \alpha, y=\sin \alpha$, the second equation is satisfied, and the first equation can be written as $\cos ^{15} 3 \alpha+\sin ^{16} 3 \alpha=1$. From the chain

$$
1=\cos ^{15} 3 \alpha+\sin ^{16} 3 \alpha \leqslant \cos ^{2} 3 \alpha+\sin ^{2} 3 \alpha=1
$$

it follows that $\cos ^{15} 3 \alpha=\cos ^{2} 3 \alpha$ and $\sin ^{16} 3 \alpha=\sin ^{2} 3 \alpha$. Therefore, either $3 \alpha=\pi / 2+\pi k$ or $3 \alpha=2 \pi k$, which means $\alpha=\frac{\pi}{6}+\frac{\pi k}{3}$ or $\alpha=\frac{2 \pi n}{3}$. This results in 9 pairs of solutions $(x, y)$:

$$
\left(\frac{\sqrt{3}}{2} ; \pm \frac{1}{2}\right) ;\left(-\frac{\sqrt{3}}{2} ; \pm \frac{1}{2}\right) ;\left(-\frac{1}{2} ; \pm \frac{\sqrt{3}}{2}\right) ;(0 ; \pm 1) ;(1 ; 0)
$$
Answer: 9.

Answer to option: $6-2$ : 9.

The equation $\cos 3 x=0$ is equivalent to

$$
3 x=\pi / 2+2 \pi k, k \in \mathbb{Z}, \quad 3 x=-\pi / 2+2 \pi n, n \in \mathbb{Z}
$$

Answers and solutions to option $\mathbf{7}-1$