Dear Captainjack2000, I wish I could do some help.
The state of a bounded particle around the nucleu is determind by ( or expressed in ) four quantum numbers:
main quantum number: n
angular magnitude quantum number: L
angular direction quantum number: m
spin quantum number: s
As to a determined $n$, $L$ can be $0, 1, 2, ..., n-2, n-1$, and there are $n$ possible values of $L$. Recall that $n >= L+1$.
For each possible value of $L$, there are $2L+1$ possible values of $m$: $-L, -L+1, -L+2, ..., 0, 1, 2, ..., L$. Recall that $L >= |m|$.
For a bounded particle around the nucleu, with spin angular momentum $s$, orbital angular momentum $L$, the allowed values of the total angular momentum quantum number $j$ will be : $j = L+s, L+s-1, L+s-2, ..., |L-s|$.
Attention that the $L$ $s$ and $j$ are always positive, while $m$ can be negative. So the lower boundary of the total angular momentum quantum numbe $j$ is set to be the absolution of $L-s$, because $s$ can be larger than $L$ sometimes.
Now, let's get down to your questions concretely.
1. s=3/2, L=2, hence j={do it yourself}.
2. Does $s$ relates to $m$ ?--- Probaly not.
$ n, L, m $ are introduced in Schrodinger's non-relativistic approach, the wave function of the bounded particle depends on $ n, L, m $ only. $s$ comes into being naturally only via Dirac's relativistic approach, although in fact $m$ was introduced phenominologically before Dirac's work.
$L$ depends on $n$, $m$ depends on $L$ (or also $n$ !). $n, L, s$ live together in bounded states, yet $s$ is insintric charactars of elementary particles whenever they are bounded or free. And for a certain particle, $s$ is determined and could only has one value. Most particles have $s$ of $\sqrt{3}/2$, with the third component $+-1/2$ (the 3/2 you referred to is the third component of the total s of some special particle). Thus, $s$ depends on none of $ n, L, m $. Moreover, no mathematical relations between $s$ and $m$, too.
