Solution:

a. Federica wins if and only if $n$ is a multiple of $3$.
If $n$ is a multiple of $3$ Federica can win: it suffices for her to carry out the move by which she removes three paperclips from the desk exactly $n / 3$ times.
On the other hand, if at some point the number of paperclips on Federica's desk is not a multiple of $3$, all the moves available to Federica leave on the desk a number of paperclips that is again not a multiple of $3$: if $k$ is even but not a multiple of $3$, then $k / 2$ cannot be a multiple of $3$ either; on the other hand, in the same way, if $k$ is not a multiple of $3$, then $k-3$ is not either. Consequently, if Federica begins the game with a number of paperclips that is not a multiple of $3$, any sequence of moves will lead to a number of paperclips that is also not a multiple of $3$; Federica therefore has no way to remove all the paperclips, since $0$ is a multiple of $3$.
In conclusion, Federica manages to win if $n$ is a multiple of $3$ between $1$ and $2015$: there are $671$ possible values for $n$.

b. This time Federica manages to win if and only if $n$ is not a multiple of $3$.
If $n$ is not a multiple of $3$ we can write it as $3k + r$, where $r$ is the remainder of the division by $3$, and hence is $1$ or $2$. By applying the first move $k$ times, Federica obtains $1$ (in which case she has won) or $2$ (in which case she uses the second move and wins).
It remains to show that if $n$ is a multiple of $3$, then Federica will never reach $1$. It suffices to show that if Federica applies a move to a number that is a multiple of $3$ she obtains a number that is a multiple of $3$ (and hence in particular she cannot reach $1$). This is true, because if $a = 3k$ is a multiple of $3$, then so are $a-3 = 3(k-1)$ and $\frac{a}{2} = 3 \frac{k}{2}$.
So Federica manages to win for $2015 - 671 = 1344$ values of $n$.