You should be aware that many people will not open "word" files! (I probably shouldn't myself but I have very strong virus protection.)
Problem 3 asks you to use Green's Theorem to show that
[tex]\int_C F(z,z*)dz= 2i \int_R \frac{\partial}{\partial z*}F(z,z*) dxdy[/tex]
where [itex]F(z,z*)= P(x,y)+ iQ(x,y)[/itex] and I presume you know that with z= x+ iy, z*= x- iy.
Green's Theorem says that
[tex]\int_C (Ldx+ Mdy)= \int_R\int \left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}\right)dxdy[/tex]
Again, the left integral is over a closed path while the right integral is over the area inside the path.
Looks to me like you need to determine what [tex]\frac{\partial}{\partial z*}F(z,z*)[/tex] looks like in terms of partial derivatives with respect to x and y. With z= x- iy, that's an exercise in using the chain rule.
Problem 4 asks you to integrate, using "Cauchy's integral theorem (NOT Cauchy's integral formula)"
[tex]\int_C \frac{g(z)}{(z-z_0)^3} dz[/tex]
where C is a closed path enclosing z0 and g(z) is analytic and single valued inside and on C.
Of course, "Cauchy's integral formula" would give the result trivially. I presume this is an exercise in proving Cauchy's integral formula. Cauchy's integral theorem says that
[tex]\int_C f(z)dz= 0[/tex] where f(z) is analytic and single valued at every point inside and on C. You can't use it directly because, of course, g(z)/(z-z0)2 is NOT analytic at z0. You might try this: since g(z) is analytic at z0, it is equal to its Taylor series there: g(z)= g(z0)+ g'(z0)(z-z0)+ (g"(z0)/2)(z-z0)(2+ ... Dividing that by (z-z0)2 gives the "Laurent" series
[tex]\frac{g(z)}{(z-z_0)^2}= \frac{g(z_0)}{(z-z_0)^2}+ \frac{g'(z_0)}{(z-z_0)}+ \frac{g"(z_0)}{2}+ ...[/tex]
You should be able to show that the integral of a constant time (z- z0)n around a circle centered on z0 (use the formula z-z0= Reit with [itex]0\le t\le 2\pi[/itex] on the circle of radius R, center z0) is 0 for every n except -1. And you should be able to get a specific value for that case.
You will need to use Cauchy's integral theorem to argue that integral around any such contour C is equal to the integral around a small circle with center z0. From your contour draw a straight line to distance R from z0, go in a circle around z0, then a straight line back to the contour. Those, patched together, give you a contour that does NOT enclose z0 and so, by Cauchy's integral theorem, has integral 0. Now, move the two straight contours together so they cancel out.