Maybe some are unfamiliar with your notation like I was. Ok, let's assume it's what I think it is: $\displaystyle \partial B_2(i)$ means the boundary of a circular disc with center at $\displaystyle i$ and radius $\displaystyle 2$. Just never seen that notation before. If so then I'll write it as:

$\displaystyle \mathop\oint\limits_{D(i,2)}\frac{e^{iz}}{z^4-4z^4}dz$

(using $\displaystyle \zeta$ makes it just unnecessarily cluttered)

So that contour encloses only the pole at the origin and you want to solve it using Cauchy's Integral Formula:

$\displaystyle f^{(n)}(w)=\frac{n!}{2\pi i}\mathop\oint\frac{f(z)}{(z-w)^{(n+1)}}dz$

where $\displaystyle f(z)$ is analytic in an on the contour. Then:

$\displaystyle \mathop\oint\limits_{D(i,2)}\frac{e^{iz}}{z^4-4z^4}dz=\oint\frac{\frac{e^{iz}}{(z^2-4)}}{(z-0)^2}dz=2\pi i\frac{d}{dz}\left(\frac{e^{iz}}{z^2-4}\right)\Bigg|_{z=0}=\pi/2$

Here's a check in Mathematica:

Code:

In[3]:=
NIntegrate[(Exp[I*z]/(z^4 - 4*z^2))*2*I*
Exp[I*t] /. z -> I + 2*Exp[I*t],
{t, 0, 2*Pi}]
N[2*Pi*I*D[Exp[I*z]/(z^2 - 4), z] /. z -> 0]
Out[3]=
1.5707963267949334 - 8.459899447643693*^-14*I
Out[4]=
1.5707963267948966

Do the other ones just like that but I bet you figured it out already.