we use the parameterization for . We can denote by the function we have to compute the integral. We have because .
For the first problem, , and so
The only pole of the integrand comes at the origin, and the residue is obviously 6 (being the coefficient of ), so you should find it easy to use the residue theorem to get the value of the integral.
Do the second problem in the same way. You should find that
(after a bit of simplification). There are two poles inside the unit circle, at . Find the residue at each of them, and use the residue theorem to get the value of the integral.