jmc

Start with small exponents and look for a pattern. We have $5^1\equiv 5(\mathrm{mod}8)$ and $5^2\equiv 1(\mathrm{mod}8)$. We can multiply both sides of $5^2\equiv 1(\mathrm{mod}8)$ by 5 to find that $5^3\equiv 5(\mathrm{mod}8)$. Multiplying both sides by 5 again, we find $5^4\equiv 1(\mathrm{mod}8)$. We see that every odd power of 5 is congruent to 5 modulo 8, and every even power is congruent to 1 modulo 8. Therefore, $5^{137}$ leaves a remainder of $\boxed{5}$ when divided by 8.