jmc

Note that for $n\ge 2$, the last two digits of $5^n$ are 25. To prove this, note that $5^2\equiv 25(\mathrm{mod}100)$, and whenever $5^{n-1}\equiv 25(\mathrm{mod}100)$, we also have $5^n=5\cdot 5^{n-1}\equiv 5\cdot 25\equiv 125\equiv 25(\mathrm{mod}100)$. Thus the tens digit of $5^{2005}$ is $\boxed{2}$.