jmc

$1995=5\cdot 399=3\cdot 5\cdot 133=3\cdot 5\cdot 7\cdot 19$. Since $3\cdot 5\cdot 7=105$ has three digits, in any expression of $1995$ as the product of two two-digit numbers, $19$ must be a proper factor of one of them. $19\cdot 3=57$ and $19\cdot 5=95$ are two-digit numbers that are divisible by $19$ and divide $1995$, but $19\cdot 7=133$ and $19\cdot 3\cdot 5=285$ are three digits, so the only possible expressions of $1995$ as the product of two two-digit numbers are $57\cdot 35$ and $95\cdot 21$. So, there are $\boxed{2}$ such factorizations.