solving one of the 5 options would be much appreciated as this will give me an idea on how to solve the rest.
Let $\gamma(w,R)$ denote the circular contour $t\mapsto w+Re^{it}$ where $0\lt t\lt2\pi$. Evaluate $$\int_\gamma\dfrac1{1+z^2}dz$$ when $\gamma$ is: $$\gamma(1,1),\quad\gamma(i,1),\quad\gamma(-i,1),\quad\gamma(0,2),\quad\gamma(3i,\pi).$$
The hard way
For example, and by definion (without Cauchy's theorems, residues and stuff), taking $\;\gamma:=\gamma(-i,1)\;$
$$z \in\gamma\implies z=-i+e^{it}\;,\;\;t\in [0,2\pi)\;,\;\;dz=ie^{it}dt$$
and the integral thus is
$$\oint_\gamma\frac1{1+z^2}dz=\int_0^{2\pi}\frac{i\;e^{it}\;dt}{1+(-i+e^{it})^2}=\left.\arctan(-i+e^{it})\right|_0^{2\pi}=$$
$$=\left.\frac12i\;\text{Log}\frac{1-(-i+e^{it})}{1+(-1+e^{it})}\right|_0^{2\pi}=\frac12i\left(\log\left|\frac{i}{i}\right|+i\arg 1-\log|1|-i\arg1+2\pi ik\right)=$$
$$=-\pi k\;,\;\;\;k\in\Bbb Z$$
The easy way:
Using Cauchy's Theorems, and since $\;\frac1{z-i}\;$ analytic on and within $\;\gamma\;$:
$$\oint_\gamma\frac{dz}{1+z^2}=\oint_\gamma\frac{\frac1{z-i}}{z+i}dz=\left.2\pi i\frac1{z-i}\right|_{z=-i}=2\pi i\frac1{-2i}=-\pi$$
