1. Solve the system:

$$
\begin{aligned}
x^{2}-y^{2} & =16 \\
3 x y^{2}+x^{3} & =260
\end{aligned}
$$
1. From the first equation, express $y^{2}=x^{2}-16$ and substitute into the second, which gives $3 x\left(x^{2}-16\right)+x^{3}=260$, reducing to $x^{3}-12 x-65=0$. By direct verification, we find that one solution is $x=5$. Dividing the polynomial $x^{3}-$ $12 x-65$ by $x-5$, we get $x^{3}-12 x-65=(x-5)\left(x^{2}+5 x+13\right)$, and since the expression in the second set of parentheses has a negative discriminant $\left(5^{2}-4 \cdot 13=-27\right)$, we conclude that $x=5$ is the only solution. Substituting back into the original system, we get $y^{2}=5^{2}-16=9$, i.e., $y= \pm 3$, so the system has two solutions: $(x, y)=(5,3)$ and $(x, y)=(5,-3)$.