jmc

Let's find the cycle of the last two digits of $7^n$, starting with $n=1$ : $07,49,43,01,07,49,43,01,\dots$ . The cycle of the last two digits of $7^n$ is 4 numbers long: 07, 49, 43, 01. Thus, to find the tens digit of $7^n$ for any positive $n$, we must find the remainder, $R$, when $n$ is divided by 4 ($R=0$ or 1 corresponds to the tens digit 0, and $R=2$ or 3 corresponds to the units digit 4). Since $2005÷4=501R1$, the tens digit of $7^{2005}$ is $\boxed{0}$.