Solution: We recall that if $z=z_0$ is an isolated singularity of a function $f$, then we have the Laurent series of $f$ in a deleted neighborhood of $z_0$, given by \[ f(z) = \sum_{n=-\infty }^{\infty } a_n z(z-z_0)^n. \] We call $a_{-1}$ the residue of $f$ at $z_0$. We also recall that \[ \int_\gamma f(z) = 2\pi \iota a_{-1}, \] where $\gamma $ is counterclockwise simple closed curve containing $z_0$ and does not contain any other poles of $f$.

We also have the Cauchy's integral theorem which says that

We also now by the Cauchy theorem, that the integration of a analytic function on a closed curve is zero. So using these facts, we now compute the integral.

Note that \begin{align*} ze^{\frac{1}{z}} & = z \left( 1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3!z^3} + \dots \right) \\ & = z + 1 + \frac{1}{2z} + \frac{1}{6z^2} + \dots. \end{align*} Therefore, the residue of $ze^{\frac{1}{z}}$ at $z=0$ will be $\frac{1}{2}$. Hence, \[ \int _C ze^{\frac{1}{z}} \mathrm{d} z = 2\pi \iota \frac{1}{2} = \pi \iota . \]

Now look at the second term under the integral. Note that \[ \tan \frac{z}{2} = \frac{\sin \frac{z}{2}}{\cos \frac{z}{2}}. \] Since $\cos \frac{z}{2}$ vanishes on $z= (2n+1)\pi$, where $n\in \mathbb{Z} $ and these points are lying outside the given circle. Therefore, $\tan \frac{z}{2}$ is analytic inside $C$ and hence, using the Cauchy's integral theorem we conclude that \[ \int_C \tan \frac{ z}{2} \mathrm{d} z = 0. \]

The final integral can be solved either using the partial fraction or by Cauchy's integral formula. Note that \begin{align*} \frac{1}{(z-1)(z-2)} & = \frac{1}{z-2} - \frac{1}{z-1}. \end{align*} The function $\frac{1}{z-2}$ is analytic inside $|z|=\frac{3}{2}$ and hence the integral over $C$ will vanish. The other integral will be \[ \int_C \frac{1}{z-1}~\mathrm{d} z = 2\pi \iota \implies \int_C \frac{1}{(z-1)(z-2)}\mathrm{d} z = -2\pi \iota . \] Thus, \begin{align*} \frac{2}{\pi\iota} \int _C \left( ze^{\frac{1}{z}} + \tan \frac{z}{2} + \frac{1}{(z-1)(z-2)} \right)\mathrm{d}z = \frac{2}{\pi \iota } \left( \pi \iota + 0 - 2\pi \iota \right) = -2. \end{align*}