Since it is a simple pole, then the residue of $\cot(z)$ at $z=0$ can be determined as $$\lim_{z\to 0}z\cot(z).$$ (Hint: As $z\to 0,$ $\frac{\sin z}z\to\, ??$)
The value thus determined will be the coefficient of $z^{-1}$ in the Laurent series of $\cot(z)$ in the annulus $0<|z|<\pi$. Can you take it from here?
In answer to your other question: yes. If it were a pole, then $|f(z)|\to\infty$ as $z\to 0,$ and if essential, then $\lim_{z\to 0}|f(z)|$ fails to exist in any sense.
There are a lot of quick tricks to figure out the location and nature of singularity without ever having to deal with Laurent expansion.
Method 1: determine how singularity transform through composition and arithmetic.
Consider 2 analytic non constant function $f,g$. Assume that you already determine zeros and isolated singularity and that you know (somehow) that $f$ have no essential singularity. For pole and zero, assume you know their order as well. Define a pole of order $k$ to be a zero of order $-k$. Every other normal point is assumed to be zeros of order $0$. Then:
$f\pm g$. Singularity point only appear at singularity of at least 1 of them. If $g$ have essential singularity the sum is still essential singularity. If they have different zeros order, then the lower one is the zero order of the singularity. For example: $\frac{1}{e^{x^{2}}-1}+\frac{1}{z}$. $\frac{1}{e^{x^{2}}-1}$ have pole order 2 (so zero order $-2$) and $\frac{1}{z}$ have zero order $-1$, so the sum have zero order $-2$ ie. pole order $-2$. If the zero order is the same, this won't work.
$fg$. Singularity point only appear at singularity of at least 1 of them. If $g$ have essential singularity, this is still essential singularity. Otherwise, simply take the sum of the order. For example, $z\frac{1}{e^{x^{2}}-1}$. $z$ have zero order $1$ and $\frac{1}{e^{x^{2}}-1}$ have zero order $-2$ so this product is a zero of order $-1$ (pole of order 1).
$\frac{f}{g}$. Singularity point only appear at singularity of at least 1 of them. If $g$ have essential singularity, this is still essential singularity. Otherwise, take zero order of $f$ minus zero order of $g$.
$f\circ g$. Singularity appear only where $g$ have one, or at preimage through $g$ of singularity of $f$. If $g$ is essential singularity then the composition might produce non-isolated singularity (so yeah it can be pretty bad). If $g$ have actual zero (zero with positive order), then the zero order of the composition is just the product of the order. If $g$ have actual pole (negative zero order), then look for the power of the highest power term in Laurent series of $f$, and multiply it with zero order of $g$ to get the zero order of the composition (if $f$ do not have highest power term, this will produce essential singularity). This is not as simple if $g$ have neither pole nor zero.
We just assumed at least 1 function do not have essential singularity. If both function have essential singularity, these trick do not work, except for composition: composition will produce essential singularity.
(a quick-and-dirty rule of thumb for remembering the above tricks is that essential singularity is infectious but harder to come up, pole and zero dominated by pole when add, cancel when multiply, and multiply when composed (for zero only))
Method 2: look up tables for information on well known function.
In particular, for polynomial, they have no singularity, and zero are the root, with zero order being multiplicity of the root.
Method 3: construct a sequence approaching the singularity that satisfy certain property:
Consider an analytic $f$. If you figure out that a singularity is at $z$ and you can construct a sequence $z_{n}\rightarrow z$ such that the sequence $f(z_{n})$ have the property:
-The modulus oscillate: then must be essential singularity.
-Do not converge: then cannot be removable singularity.
-Do not diverge to infinity: then cannot be pole.
Unfortunately, you cannot rule out essential singularity with this trick.
Now back to example in your case. We can apply these tricks.
$\frac{\sin z}{z^{2}}$. Look up to find $\sin z$ have zero order $1$. $z^{2}$ have zero order $2$. So this is pole order $1$ ($1-2=-1$).
$\frac{\cos z-1}{z^{3}}$. Unfortunately, the trick won't help you to find the zero of the numerator. Simply look up Taylor series expansion to find $\cos z-1$ have zero order $2$. $x^{3}$ have zero order $3$. So this become pole order $1$.
$z^{4}\sin\frac{1}{z}$. $\frac{1}{z}$ is pole order $1$, but $\sin$ have infinite Taylor expansion, so $\sin\frac{1}{z}$ is essential singularity. $z^{4}$ do not have essential singularity, so the product is still essential singularity.
$\frac{1+z}{1-z^{4}}$. $1-z^{4}$ have zero order $1$ at these $4$ point. $1+z$ have zero order $1$ at only $z=-1$. So at $z=-1$ this cancel, so is not a pole. At other point there are no zero order in the numerator to cancel out that in the denumerator, so they are pole of order $1$.
Best Answer
Since $\sin\left(\frac\pi z\right)$ has a simple zero at $z=1$ and $\cos\left(\frac\pi z\right)$ has not zero there,$$\frac{\sin\left(\frac\pi z\right)}{(z-1)\cos\left(\frac\pi z\right)}$$has a removable singularity at $z=1$ (the simple zero in the numerator cancels the simple zero in the denominator). It's as simple as that.