Solution:

a. Since $2012$ is a multiple of $4$, we can consider the quadruples of consecutive numbers $(k, k+1, k+2, k+3)$ and observe that it is possible to return to the starting point every four steps, since it suffices to jump first to the right by $k$, then to the left by $k+1$, again to the left by $k+2$, and finally to the right by $k+3$:
$$k-(k+1)-(k+2)+k+3=0$$
Obviously one can swap the jumps to the right and to the left and permute their order.

b. If we consider the numbers from $1$ to $2013$, we have that, regardless of the sign we place in front of each number, the sum will be odd, since there are $1007$ odd terms, and therefore it cannot be $0$.

c. It turns out that it is possible to return to the starting point for all numbers that have remainder $0$ or $3$ upon division by $4$. For multiples of $4$ one can use the same reasoning used for $2012$. For numbers belonging to the residue class $3$ modulo $4$, we observe that with the first $3$ jumps the flea can return to the origin: $1+2-3=0$, after which there remain a number of jumps that is a multiple of $4$, so they can be grouped as shown for the case $n=2012$. Alternatively, one can reduce to the previous case by introducing a virtual jump of length $0$ and grouping $4$ at a time.
For $n$ that gives remainder $1$ or $2$ upon division by $4$ it is not possible to return to the starting point. Indeed, however the direction of the jumps is chosen, that is, the signs, we will have a sum with an odd number of odd terms, hence odd itself, and, in particular, different from $0$.