1) The poles of $f$ are the zeroes of $1/f$ and vice versa.
2) $f(z)=\sin(1/z)$ is defined and holomorphic everywhere except $z=0$, where it has an essential singularity. Zeroes when $1/z=\pi n i \implies z=-i/\pi n$
3) $f$ takes all complex values near each of $\{0, \infty \}$, and near both has infinitely many zeroes. Not sure exactly what they mean by "nature" though.
If you write down the Laurent series carefully, I suppose these answers will come like A,B,C to you.
Since $f$ has a pole of order $m$, it follows that $f(z)=(z-z_0)^{-m}h(z)$, where $h(z)$ is an analytic function on the domain where $f$ is analytic, and the point $z_0$. Now, we can write down $\frac{f'(z)}{f(z)}$ as:
$$
\frac{f'(z)}{f(z)} = \frac{(-m)(z-z_0)^{-m-1}h(z) + h'(z)(z-z_0)^{-m}}{(z-z_0)^{-m}h(z)}
$$
There is only one factor at the bottom of $(z-z_0)$, that is to say,
$$
\frac{f'(z)}{f(z)} = \frac{(-m)}{(z-z_0)} + \frac{h'(z)}{h(z)}
$$
The first function has a pole at $z_0$ of order $1$, and the second one doesn't have a pole at $z_0$ because $h(z_0) \neq 0$. Hence the order of $z_0$ as a pole of $f$ is $1$.
When we multiply fractions, we add the pole factors. To see this, suppose $f$ and $g$ have poles at $z_0$ or orders $m$ and $n$ respectively. Now, note that by definition, $f(z)=(z-z_0)^{-m}f_1(z)$ and $g(z)=(z-z_0)^{-n}g_1(z)$ where $f_1$ and $g_1$ are holomorphic at $z_0$ and don't vanish at $z_0$. Just multiply these to see that $fg(z)=(z-z_0)^{-(m+n)}f_1g_1(z)$, where $f_1g_1$ is holomorphic at $z_0$ and doesn't vanish at $z_0$. This tells us the order of $z_0$ as a pole of $fg$ is $-(m+n)$.
For dividing functions, see that you get $m-n$ when $m>n$, and the pole vanishes if $m \leq n$.
Actually, when you Laurent expand $f$ around $z_0$, the negative coefficient $c_m$ will be non-zero, but all coefficients of terms less than $-m$ will all be zero. When you differentiate this relation and get the term $f'$ then see what the result is for yourself about the order of $z_0$ as a pole of $f'$. You will see that the order actually increases by $1$, it would now be $m+1$.
Best Answer
Because $\sin x$ has a zero of order 1 at $\pi$, you would end up with a removable singularity at $\pi$.