Let $f(z)$ be
$f(z) = \frac{2}{(z-1)^2} +\frac{3}{(z-1)}+\frac{\sin z}{e^{z}}$ ,
calculate
$I = \int_{|z|=5} f(z) dz$
I already use $\int \frac{dz}{(z-a)^n}=0$ and $\int \frac{dz}{z-a}=2 \pi i$
to solve $\frac{2}{(z-1)^2}$ and $\frac{3}{(z-1)}$ and know the answer are $0$ and $6 \pi i $,But I don't know how to calculate $\frac{\sin z}{e^{z}}$.
How to calculate $\frac{\sin z}{e^{z}}$??
You can choose to use the reidue theorem:
$$\int_\gamma f(z) dz = 2i \pi \sum_{k=1}^n \mathrm{Res}(f, z_k)$$
In the circle defined by $|z| = 5$, you must look for the singularities you have. As your function is as sum of 3 other functions, we can apply the residue theorem to each of them separatly and then sum the results.
For the first term: double pole at $z = 1$ (double because of the "$^2$"). Is it in the circle $|z| = 5$ ? Yes. Same idea for the second term: simple pole at the same point.
About the third term, let's look for the singularities:
$$e^z = 0 = \cos(z) + i \sin(z)$$
Their is not value of $z$ such as the equation above is satisfied. So the residue of this term is equal to $0$.
Note that I used an alternative method, and that yours (first Cauchy formula) also works. If you are not familiar with the residue theorem, use the Cauchy Goursat theorem which is much easier to understand. It stands that for any analytical functions in your domain (here, $|z| \le 5$), its definite integral along your curve $\gamma$ will be equal to $0$.