1. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
Answer 36.

Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, otherwise the product will be at least 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, or 47 - 14 options, for 3 we get: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 - 10 options, for 5: 3, 5, 7, 11, 13, 17, 19 - 7 options, and for 7: 3, 5, 7, 11, 13 - 5 options. In total, $14+10+7+5=36$ options.