For $1$, what I'd do is ignore the $\cos z$ term since $\cos$ is an entire function so its integrals around closed curves vanish; then write the other term's Laurent series
$-z(1-1/2z^2+...)=-z+1/2z-...$ to get that the residue at $0$ is $1/2$. Clearly there are no other poles.
For question $2$, note that on the GRE it's not important to be able to show very rigorously that the series converges; but note that it increases for $x_n<\sqrt{2}$ and decreases for $x_n>\sqrt{2}$ (this is taking $x_1>0$ and leaving the negative case to you) so it at least won't diverge to infinity. Then to find the limit $L$ assuming it exists, in this sort of problem one takes the limit of both sides and writes $L=1/2(L+2/L)$, then solves the resulting quadratic. As I've already hinted, the positive solution is $\sqrt{2}$.
I can't think of a very good way to do question $3$ quickly. Rewriting into rectangular coordinates and doing a grotesque implicit differentiation would take rather more than the $3$ minutes I aim for on GRE problems. But it's multiple choice, and we can at least see that we get up to $y=1/\sqrt{2}$ at $\theta=\pi/6$ and by $\theta=\pi/4$ we're down to $0$.
As to the final question, remember that only one-to-one functions can have inverses, and that the inverses are also one-to-one. That is, there is no such $f$.