You should be aware that a simple pole is a point in which the Laurent series of $f$ is
$$
f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\ldots
$$
The residual is $a_{-1}$, the coefficient of $1/(z-z_0)$.
Now, we have
$$
(z-z_0)f(z)=a_{-1}+a_0(z-z_0)+a_1(z-z_0)^3+a_2(z-z_0)^3+\ldots
$$
that gives $a_{-1}$ when you do the limit $z\to z_0$.
When $f$ is a rational function, ratio of two polynomials $p$ and $q$, with a single pole, $q$ should be of the form
$$
q(z)=(z-z_0)Q_1(z)
$$
with $Q_1(z_0)\neq0$, and
$$
q'(z)=Q_1(z)+(z-z_0)Q_1'(z)
$$
so that the limit for $z\to z_0$ is $q'(z_0)=Q_1(z_0)$.
At this point
$$
(z-z_0)f(z)=(z-z_0)\frac{p(z)}{(z-z_0)Q_1(z)}=\frac{p(z)}{Q_1(z)}
$$
and for $z\to z_0$ we have
$$
\frac{p(z_0)}{Q_1(z_0)}=\frac{p(z_0)}{q'(z_0)}
$$
I add that you could use the same formula for each single pole, also if the function has more than one single pole.
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
In order to determine the Laurent expansion at $1$ and $-1$ see Claude Leibovici's answer.
To evaluate the residues, you may also note that $-1$ and $+1$ are simple poles. Therefore $$\mbox{Res}\left(\frac{\sinh(z)/z^4}{1-z^2},\pm 1\right)=\left.\frac{\sinh(z)/z^4}{(1-z^2)'}\right|_{z=\pm 1} =\left.\frac{\sinh(z)/z^4}{-2z}\right|_{z=\pm 1}=-\frac{\sinh(1)}{2}.$$ See the detailed explanation here.