1. Given a two-digit number, where the first digit is greater than the second. Its digits were rearranged, and a new two-digit number was obtained. The difference between the first and the second number turned out to be a perfect square. How many possible two-digit numbers satisfy this condition?
1. Answer: 13 numbers.

Let the first number be $10a + b$. Then the second number is $10b + a$, where $a > b > 0$. The difference is $9(a - b)$, which implies that the difference $a - b$ is a perfect square. The difference can be 1 or 4. For the first case, we have 8 values for the larger number $(21, 32, \ldots, 98)$, when the difference is $3^2$, and for the second case, there are 5 values $(51, 62, 73, 84, 95)$, when the difference is $6^2$. In total, 13 numbers.