1. Find all pairs of natural numbers $a, b$ greater than 1 such that both their sum and product are powers of prime numbers (with positive integer exponents).
1. From the condition for the product, it follows that $a$ and $b$ are powers of the same prime number $p: a=p^{r}$, $b=p^{s}$, where $r, s$ are positive integers. If $p$ were odd, the sum $a+b$ would be divisible not only by the number $p$ but also by the number 2, so it would not be a power of a prime number. If $p=2$ and $r \leq s$. Therefore, the only remaining possibility is: $a=b=2^{r}$, where $r$ is a positive integer. The check $a+b=2^{r}+2^{r}=2^{r+1}$ and $a b=2^{2 r}$ confirms that the solutions are all pairs $(a, b)=\left(2^{r}, 2^{r}\right)$, where $r$ is a positive integer.

For a complete solution, award 6 points. For identifying that $a, b$ are powers of the same prime number, award 1 point, and another point for justifying that $p=2$. Then, 2 points for proving $r=s$ (due to symmetry, one can immediately assume $r \leqq s$), and 1 point for the check.