An important point is that a "pole" is actually the same thing as a removable singularity, if we think of our function as a map that takes values on the Riemann sphere (which is the complex plane with a point at $\infty$ added; the complex structure near $\infty$ comes from the map $z\mapsto 1/z$).
So a function that has a removable singularity or a pole at $z_0$ doesn't have a "real" singularity there at all; rather, we can extend the function to an analytic or meromorphic function in $z_0$. If we cannot extend the function in this way, the singularity is indeed "essential"; i.e., we cannot get rid of it. Thus the terminology is not one that is merely used for convenience or pedagogical purposes; rather, it is extremely natural.
As has already been mentioned, the magic of complex numbers results in many beautiful facts about essential singularities: functions with these singularities are very far from extending continuously.
The simplest of these facts is the Casorati-Weierstraß Theorem: The image of a neighborhood of an essential singularity is dense in the complex plane.
This is just a consequence of the removable singularities theorem. (If $f$ omitted a neighborhood of $a$, we could postcompose $f$ with a Möbius transformation that takes $a$ to infinity and see that the resulting function has a removable singularity.)
The most well-known result of this type is Picard's theorem which was already mentioned.
There are various beautiful strengthenings of Picard's theorem that arise from Nevanlinna theory, and Ahlfors's theory of covering surfaces.
So all essential singularities have some things in common, but on the other hand this should not lead us to believe that they are all the same. What they have in common is complicated behaviour, but they can be complicated in very different ways! Indeed, different transcendental entire functions (those that have an essential singularity at infinity; i.e. are not polynomials) can vary very much with respect to their behavior near infinity. Just for example, for some such functions, such as $z\mapsto e^z$, there exist curves tending to infinity on which the function is bounded, while for others this is not the case.
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).
Best Answer
Singularities are the points $z$ such that $\cos (1/z)=0, $ which occurs when $1/z=\pi/2+n\pi$ so that you can write the values of $z$ now, where $n$ is any integer.
Second part: We have Tylor series expansion for $\sec z$ by which we can get the Laurent series expansion for $1/\cos (1/z)$ with infinitely many terms with negative powers of $z$. It is very well explained over here Singularities of $ {1}/{\cos(\frac{1}{z})}$
Therefore Sine function after composition has Laurent series expansion at $z=0$ with infinitely many terms having negative powers of $z.$ Hence the singularities are of essential type.