Answer
a) $l$ can be $0$, $1$, $2$ or $3$;
b)
1) $3$
2) $3$
3) $1$
4) $1$
Work Step by Step
a) Angular quantum number has to be between $0$ and $n-1$, so for $n=3$, possible options for $l$ are $0$, $1$, $2$ and $3$.
b)
1) If $n=3$ and $l=1$, allowed values for magnetic quantum number are $-1$, $0$ and $1$, so there can be only $3$ orbitals with such principal and angular quantum numbers ($3p_{x}$, $3p_{y}$ and $3p_{z}$).
2) As shown in the previous example, allowed values for magnetic quantum number are $-1$, $0$ and $1$, so there can be only $3$ orbitals with such principal and angular quantum numbers ($2p_{x}$, $2p_{y}$ and $2p_{z}$).
3) There is only one orbital with this combination of quantum numbers, and that is $3p_{x}$.
4) If $n=1$, the only value angular quantum number can take is $0$, which implies that the only possible value for magnetic quantum number is also $0$. Hence, there is only one orbital with principal quantum number equal to $1$, and that is $1s$ orbital.
