The contour $\Gamma$ is parameterised by $r:[-\pi,\pi] \rightarrow \mathbb{C}$ is given by $r(\theta)=3e^{i\theta}+1$.
(a) $\int_{\Gamma} \frac{\sin(z)}{z-1}\ dz = 2 \pi i \sin(1)$?
(b) $\int_{\Gamma} \frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-1)(z-2)}\ dz = \frac{\frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z-2)}}{(z-1)}\ dz = 2 \pi i \frac{\sin (\pi)+\cos(\pi)}{-1} = 2 \pi i$?
(c) $\int_{\Gamma} \frac{e^z}{z^2+4}\ dz = \int_{\Gamma} \frac{\frac{e^z}{z+2i}}{z-2i}\ dz = 2\pi i \frac{e^{2i}}{4i} = \frac{\pi}{2}e^{2i}$?
(d) $\int_{\Gamma} \frac{e^z}{z(z-1)(z-2)}\ dz $
Do I need partial fractions or can do this: $\int_{\Gamma} \frac{\frac{e^z}{z(z-1)}}{(z-2)}\ dz =2 \pi i \frac{e^2}{2} $
I need some clarification on my answers. Thanks.
The first one is correct, but you forgot to take into account all of the isolated singular points. Let's work through the problems one by one.
Throughout this post I will use the theorem that if we can write $f(z)=\frac{p(z)}{q(z)}$ where $p(z_0)\neq 0$, $q(z_0)=0$, and $q'(z_0)\neq0$ for some isolated singularity $z_0$, then
$$\underset{z=z_0}{\mbox{Res}} \, \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$$
A) Using the Cauchy residue theorem and identifying the isolated singular point to be $z_0=1$, we can solve the integral
$$\int_{\Gamma} \frac{\sin(z)}{z-1} dz = 2 \pi i \bigg( \underset{z=1}{\mbox{Res}} \, \frac{\sin(z)}{z-1} \bigg)=2 \pi i \big(\frac{\sin(1)}{1} \big)=2 \pi i \sin(1)$$
You got this question right because you used the appropriate equation and there was only one pole.
B) Using the Cauchy residue theorem and identifying the isolated singular points to be $z_0=1$ and $z_1=2$, we can solve the integral
$$\begin{aligned} \int_{\Gamma} \frac{\sin(z^2 \pi)+\cos(z^2 \pi)}{(z-1)(z-2)} dz =& 2 \pi i \bigg( \underset{z=1}{\mbox{Res}} \, \frac{\sin(z^2 \pi)+\cos(z^2 \pi)}{(z-1)(z-2)} + \underset{z=2}{\mbox{Res}} \, \frac{\sin(z^2 \pi)+\cos(z^2 \pi)}{(z-1)(z-2)} \bigg) \\ =& 2 \pi i \bigg(\frac{\frac{\sin(1 \pi)+\cos(1\pi)}{1-2}}{1} + \frac{\frac{\sin(4 \pi)+\cos(4\pi)}{2-1}}{1} \bigg) \\ =& 2 \pi i (1+1) \\ =& 4 \pi i\end{aligned}$$
C) Using the Cauchy residue theorem and identifying the isolated singular points to be $z_0=2i$ and $z_1=-2i$, we can solve the integral
$$\begin{aligned} \int_{\Gamma} \frac{e^z}{z^2+4} dz =& 2 \pi i \bigg( \underset{z=2i}{\mbox{Res}} \, \frac{e^z}{(z-2i)(z+2i)} + \underset{z=-2i}{\mbox{Res}} \, \frac{e^z}{(z-2i)(z+2i)} \bigg) \\ =& 2 \pi i \bigg(\frac{\frac{e^{2i}}{(2i+2i)}}{1} + \frac{\frac{e^{-2i}}{(-2i-2i)}}{1} \bigg) \\ =& 2 \pi i \big(\frac{e^{2i}}{4i} + \frac{e^{-2i}}{-4i} \big) \\ =& \pi i \sin(2)\end{aligned}$$
Note at the end we used the identity $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$.
D) Using the Cauchy residue theorem and identifying the isolated singular points to be $z_0=0$, $z_1=1$ and $z_2=2$, we can solve the integral
$$\begin{aligned} \int_{\Gamma} \frac{e^z}{z(z-1)(z-2)} dz =& 2 \pi i \bigg( \underset{z=0}{\mbox{Res}} \, \frac{e^z}{z(z-1)(z-2)} + \underset{z=1}{\mbox{Res}} \, \frac{e^z}{z(z-1)(z-2)} + \underset{z=2}{\mbox{Res}} \, \frac{e^z}{z(z-1)(z-2)} \bigg) \\ =& 2 \pi i \bigg(\frac{\frac{e^{0}}{(0-1)(0-2)}}{1} + \frac{\frac{e^{1}}{(1)(1-2)}}{1} + \frac{\frac{e^{2}}{(2)(2-1)}}{1} \bigg) \\ =& 2 \pi i \big(\frac{1}{2} + \frac{e}{-1} + \frac{e^2}{2} \big) \\ =& \pi i (e-1)^2 \end{aligned}$$
It seems from these questions that when using the Cauchy residue theorem
$$\int_{C} f(z) \, dz = 2 \pi i \sum_{k=0}^{n} \underset{z=z_k}{\mbox{Res}} \, f(z)$$
You forgot that you had to calculate the residue for $\textit{every}$ isolated singular point inside of our contour. Since the contour $\Gamma$ is a circle of radius $3$ centered at $z=1$, it contains all of the isolated singularities for each function in parts A) through D).