Okay, it looks mostly good, except for a couple things. First of all, in the a0 calculation, I'm not understanding where you said:
\displaystyle \frac{1}{2L} \int_{-L}^{L} f(x) dx = \frac{1}{4L} \int_{-L}^{0} f(x) dx + \frac{1}{4L} \int_{0}^{L} f(x) dx
That should be \displaystyle \frac{1}{2L} \int_{-L}^{L} f(x) dx = \frac{1}{2L} \int_{-L}^{0} f(x) dx + \frac{1}{2L} \int_{0}^{L} f(x) dx
You already cut the limits of integration in half (going from -L to 0 instead of -L to L) so you don't need to cut the coefficient in half as well. It seems like that's what you did, because you did it again in the an and bn calculations. That error threw your answer off by a factor of (1/2).
The only other error I see is in the second and third last lines of the bn calculations. You're saying that cos(nx) evaluated at x=0 is 0, which is isn't, cos(0)=1. Always be careful to actually evaluate the function at the integration limits, and don't just drop the x=0 terms. They're often zero, but not always. Especially in Fourier calculations, they're very often not zero. It looks like you did the same thing with the an calculation, although it did turn out to be right since sin(0)=0.
One last thing, which I guess isn't technically an error, but I wouldn't be surprised if you lost marks for it. At the end of your an and bn calculations (and it carried through to your final answer), you ended up with some terms like cos(n*pi) and sin(n*pi). Instead of leaving them as they are, why not evaluate them, since trig functions tend to be nice at integer multiples of pi? What are the values of those two functions when n is even? What about odd? If you do that correctly, you should find that one of them drops out entirely, and the other is only non-zero when n is odd. This simplifies your answer a lot. Since you're asked for the first four non-zero terms, not doing this would actually make your final answer incorrect.
Hope that helps.
EDIT:
I noticed you mentioned that if the function was odd (I think you meant even), it would be easier to evaluate. There are tricks you can use to make a function 'nicer.' They can cut down your work quite a bit, though they might not. For example, with this function, you could shift f(x) to the left by (pi/2) and make it an even function. You could also shift it (the original function, not the even function I just mentioned) down by (1/2) and make it into an odd function. Both of those shifts will affect the Fourier series in a predictable way, so that if you can find the Fourier series for the shifted function, you can easily convert to the Fourier series of the original function. In this case, shifting the function down by (1/2) makes the calculation a heck of a lot easier. It turns your function into an odd function, so you know an is 0 for the Fourier series of the shifted function, and you only have to find a0 and bn for the shifted function. But then, a constant amplitude shift like that only changes the a0 term. So all you'll have to do to get back to the Fourier series of the original function is either add or subtract (1/2) to the value of a0 you found, and you're done! That saves a lot of work (especially for more complicated problems), and leaves less places for you to make errors.
