A perfect square is written down, but only its last two digits can still be read. Which of these pairs of digits is IMPOSSIBLE?

A square ending in 6 must come from a number ending in 4 or 6. Squaring those gives $(10a+4{)}^2=100a^2+80a+16$ and $(10a+6{)}^2=100a^2+120a+36$, so the tens digit is the last digit of $8a+1$ or of $12a+3$ — ODD in every case. Squares ending in 6 therefore always have an odd tens digit: 16, 36, 196, 256, 576, 676 and so on. Since 66 has the even tens digit 6, no square ends in it. The other three are easy to produce: $25=5^2,36=6^2$ and $64=8^2$. Checking only the units digit is what makes 66 look acceptable — 6 is a perfectly legal ending, it is the pair that fails.