Melody, you're on the right track, but here is my attempt:
You're right about the odd and even switching,
5^3=125
5^4=625
5^5=last three digits of 125
5^6=last three digits are 625
So, for odd powers of five, the last three digits are 125 and for even powers five, the last three digits are 625. By the divisibility rule for 8, and we have \(\frac{125}{8}\) since 137 is odd. Thus, the remainder is \(\boxed{5}.\)
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