Let $C$ be the simple, positively oriented circle of radius $2$ centered at the origin in the complex plane then
$$\frac{2}{\pi i}\int_c (ze^{1/z}+\tan(\frac{z}{2})+\frac{1}{(z-1)(z-3)^2})dz$$
my attempt:
since the coefficient of $z e^{1/z}$ is 2and coefficient of $\tan(\frac{z}{2})$ and $\frac{1}{(z-1)(z-3)^2}$ is 0
then $\frac{2}{\pi i}\int_c (ze^{1/z}+\tan(\frac{z}{2})+\frac{1}{(z-1)(z-3)^2})dz=2\pi i \times \frac{2}{\pi i}\times 2=8$ correct?
When you write "coefficient" you actually mean residue.
The function $$f(z):=z\,e^{1/z}= z+1+{1\over2z}+{1\over6z^2}+\ldots$$ has an essential singularity at $z=0$ with residue ${\displaystyle{1\over2}}$.
The function $z\mapsto\tan{\displaystyle{z\over2}}$ is analytic in the closed disc bounded by $C$.
The function $g(z):={\displaystyle{1\over(z-1)(z-3)^2}}$ has a first order pole at $z=1$ with residue $${1\over(z-3)^2}\Biggr|_{z=1}={1\over4}\ .$$