Determining Singularities. Couple things,
-Can someone please explain why 
f(z)=1/sin(z)  is considered a simple pole and not a removable singularity?
-Also for the function,

I know singularities exist at 2i, -2i, -3, 1. 
Which to me seems like there are double poles at 2i,-2i and simple poles at -3,1.
However the solutions sheet indicates only -2i is a double pole and the rest are simple poles. Can anyone tell my why 2i is a simple pole??

For the first question, I assume you mean at $z = 0$ (or any other zero of $\sin z$). In this case, the denominator is $\sin 0 = 0$, and this is a zero of order $1$. On the other hand, the numerator has no zero, so there is no cancellation - hence the singularity is not removable.
For the second question, note that there is cancellation. The numerator has a zero of order $1$ at $z = 2i$, while the denominator has a zero of order $2$. The net effect of this is a zero of order $-1$, or a pole of order $1$.
