Evaluate $\hat f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$ for $f(t)=\sum_{n=0}^{\infty}\frac{1}{2^n}1_{[-2,2]}(x-2n)$ 
$\hat f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$, for $f \in L^1(\mathbb R)$
Let $f(t)=\sum_{n=0}^{\infty}\frac{1}{2^n}1_{[-2,2]}(t-2n)$, where $1_{[.,.]}$ is the indicator function
I want to determine $\hat f(x).$

I think we need to interchange the series and the integral which should be no problem because
$\frac{1}{2^n}1_{[-2,2]}(x-2n) \ge 0$ so we can use Fubini.
Now, using $1_{[-2,2]}(x-2n)=1_{[2n-2,2n+2]}(x)$:
$\hat f(x)=\frac{1}{\sqrt{2\pi}}\sum_{n=0}^{\infty}\frac{1}{2^n}\int_{-\infty}^{\infty}1_{[2n-2,2n+2]}(t)e^{-itx}dt=\frac{1}{\sqrt{2\pi}}\sum_{n=0}^{\infty}\frac{1}{2^n}\frac{e^{-i(2n+2)}-e^{-i(2n-2)}}{-ix}$
Is that correct so far?
I thought about using $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$ to simplify this term but I don't know if that will help

It's correct, but at end there is no need to that formulas for simplification of sum. Just observe that all terms display in positive and negative, so annihilate each other, but these two:
$$\hat{f}(x) = \frac{1}{\sqrt{2\pi}}(\frac{e^{-2i}}{ix}+\frac{1}{2}.\frac{1}{ix})$$
