I'm not quite sure what you're asking...but...first, in the world of complex variables trigonometric functions and exponential functions all unify. Thus by moving to complexes we have a nice uniform way of dealing with sine and cosine.
If $c_n=a_n+ib_n$, then
$$ c_ne^{int}+c_{-n}e^{-int} = c_ne^{int}+\bar{c_{n}}e^{-int} =$$ $$ (a_n+ib_n)(\cos(nt)+i\sin(nt))+(a_n-ib_n)(\cos(nt)-i\sin(nt)) = 2a_n\cos(nt)-2b_n\sin(nt) $$
So dealing with trig functions or exponentials is just a matter of notation/taste.
Now to address your question as to what these series are for...
Recall that an analytic function is equal to its Taylor series:
$$ f(x) = \sum\limits_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k $$
If we stop this sum, say at $N$, we have
$$ f(x) \approx \sum\limits_{k=0}^N \frac{f^{(k)}(a)}{k!}(x-a)^k $$
which is a polynomial approximation of $f(x)$. Polynomials are nice. Taylor polynomials give us easily understood approximations of our function.
Now Fourier series do essentially the same thing for periodic functions. If we stop the summation at some $N$, we get a Fourier polynomial. This is a nice approximation made up of "easy" to understand trig functions.
From a theoretic viewpoint, Taylor series are wonderful because you can treat analytic functions sort of like polynomials. Fourier series allow one to treat nice periodic functions sort of like trig functions.
You know that
$$f(x) = \frac{a_0}{2} + \sum_{n = 1}^\infty (a_n \cos(nx) + b_n\sin(nx)), \quad x \in [-\pi,\pi].$$
So
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{a_0}{2}\int_{-\pi}^\pi g(x)\, dx + \sum_{n = 1}^\infty \left(a_n \int_{-\pi}^\pi \cos(nx)g(x)\, dx + b_n \int_{-\pi}^\pi \sin(nx)g(x)\, dx\right).$$
This step is justified by the uniform convergence assumptions given in the problem. By definition of the Fourier coefficients $\alpha_n$ and $\beta_n$, we have
$$\pi \alpha_0 = \int_{-\pi}^\pi g(x)\, dx,$$
$$\pi \alpha_n = \int_{-\pi}^\pi \cos(nx)g(x)\, dx,\quad n \ge 1$$
and
$$\pi \beta_n = \int_{-\pi}^\pi \sin(nx)g(x)\, dx,\quad n \ge 1.$$
Therefore
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{\pi a_0\alpha_0}{2} + \sum_{n = 1}^\infty (\pi a_n\alpha_n + \pi b_n \beta_n),$$
or
$$\int_{-\pi}^\pi f(x)g(x)\, dx = \frac{a_0 \alpha_0}{2} + \sum_{n = 1}^\infty (a_n \alpha_n + b_n \beta_n).$$
Best Answer
Let the Fourier Series representation for $f$ and $g$ be written respectively by
$$f(x)=\sum_{n=-\infty}^\infty \alpha_ne^{inx}$$
$$g(x)=\sum_{n=-\infty}^\infty \beta_ne^{inx}$$
with coefficients $\alpha_n$ and $\beta_n$ given by
$$\alpha_n=\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}\,dt$$
$$\beta_n=\frac1{2\pi}\int_{-\pi}^\pi g(t)e^{-int}\,dt$$
The convolution of $f$ and $g$ is defined as
$$\begin{align} (f*g)(x)&\equiv \frac1{2\pi}\int_{-\pi}^\pi f(t)g(x-t)\,dt\\\\ &=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{n=-\infty}^\infty \beta_n e^{in(x-t)}\right)\,dt\\\\ &=\sum_{n=-\infty}^\infty \beta_n e^{inx}\left(\frac1{2\pi}\,\int_{-\pi}^\pi f(t)e^{-int}\,dt\right)\\\\ &=\sum_{n=-\infty}^\infty \alpha_n\,\beta_n e^{inx} \end{align}$$
and we are done!