a) $\displaystyle{f(z)=\dfrac{1}{e^{1/z}-1}}$.
It says $f:\mathbb C\setminus\{0\}\to\mathbb C$, but this is incorrect, because $f$ has a simple pole at $z=\dfrac{1}{2\pi ki}$ for each nonzero integer $k$, and $z=0$ is not even an isolated singularity. If you change the codomain to $\mathbb C\cup\{\infty\}$ and think of $f$ as a meromorphic function, then it has an essential singularity at $0$.
In fact, you can show that $f(D(0,r)\setminus\{0\})=(\mathbb C\cup\{\infty\})\setminus\{0,-1\}$ for all $r>0$, using elementary properties of the exponential function.
b) $\displaystyle f:\mathbb{C}\backslash\{0,2\}\rightarrow\mathbb{C},\ f(z)=\frac{\sin z ^2}{z^2(z-2)}$
Evaluate $\lim\limits_{z\to 0}f(z)$ and $\lim\limits_{z\to 2}f(z)$. One is finite, the other is $\infty$, so you have a removable singularity and a pole.
c) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\cos\left(\frac{1}{z}\right)$
In this case, you should be able to show, even just using real variables, that $\lim\limits_{z\to 0}f(z)$ does not exist in either a finite or infinite sense. Sketch a graph of $y=\cos(1/t)$ close to $0$. Another useful tool is the Laurent series, which in this case is obtained from the power series expansion of $\cos$ by substitution of $1/z$. And similarly to a), you could use elementary properties of the exponential function along with the identity $\cos(z)=\frac{1}{2}(e^{iz}+e^{-iz})$ to find the image of a small punctured disk at $0$.
d) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{1-\cos\left(\frac{1}{z}\right)}$
Similarly to a), this is incorrect. Either the domain or the codomain should be changed. If you don't change the codomain, then $f$ is undefined where $\cos(1/z)=1$, and there is not an isolated singularity at $0$. If you allow meromorphic functions, then it is an essential singularity at $0$. (And again you could even explicitly find the range, or you could more simply show that no limit exists by choosing special values.)
e) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{\sin\left(\frac{1}{z}\right)}$
See a) and d).
Since for any $\,k\in\Bbb Z-\{0\}\,$ we have that
$$\lim_{z\to k\pi}(z-k\pi)\frac{z-1}{z^2\sin z}\stackrel{\text{L'Hospital}}=\lim_{z\to k\pi}\frac{z-1}{2z\sin z+z^2\cos z}=$$
$$=\frac{k\pi -1}{2k\pi \sin k\pi+k^2\pi^2\cos k\pi}=\frac{(-1)^k(k\pi -1)}{k^2\pi^2}$$
all these singularities are isolated (in fact, simple poles).
About the case $\,k=0\,$:
$$\lim_{z\to 0}z^3\frac{z-1}{z^2\sin z}=\lim_{z\to 0}\frac{z}{\sin z}(z-1)=-1$$
Thus we have here a pole of order $\,3\,$
Best Answer
I'm not sure why power series didn't work for you. At any $n\pi,\sin n\pi = 0, sin'(n\pi)=\cos n\pi =(-1)^n,$ and we don't need anything more. Why? Because this shows that near $n\pi,$ $\sin z = (-1)^n(z-n\pi)+O(|z-n\pi|^2).$ Thus $\lim_{z\to n\pi}(z-n\pi)/sin z = (-1)^n.$