Answer: 7.

Solution: Such numbers have the form $2^{k} \times p^{2}$, where $p-$ is an odd prime number. Clearly, $p$ does not exceed 7, since the result must be a two-digit number. If $p=3$, then $k=0,1,2,3$; If $p=5$, then $k=0,1$; if $p=7$, then $k=0$. In total, there are 7 options.

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5. Let's call a number "marvelous" if it has exactly 3 different odd natural divisors (and an arbitrary number of even divisors). How many "marvelous" two-digit numbers exist?