Answer: 35

Small Hint:

Write $6=7-1$ and $8=7+1$

Big Hint:

Modulo $49,$ only the constant and linear terms of each binomial expansion survive

Solution:

Modulo $49,$ every binomial term containing $7^2$ vanishes. Since $83$ is odd, $$6^{83}=(7-1)^{83}\equiv -1+83\cdot7$$ and $$8^{83}=(7+1)^{83}\equiv 1+83\cdot7.$$ Their sum is congruent to $166\cdot7=1162\equiv35\pmod{49}.$