Fourier transform of $\sum_{n=-\infty }^\infty e^{-|t-2n|}$
First of all, the signal is periodic, so its spectrum must be discrete. Hence, your final solution ("Original answer") cannot be correct.

I would go the following way:


Calculate $X(f)=\mathcal{F}\left\{\exp(-\lvert t\rvert)\right\}$
Apply the time-shifting property of the Fourier Transform to transform all time-shifts of the function.
See that you can factor out $X(f)$ from the resulting sum and then see that the terms in the sum correspond to a comb function where the diracs are $\frac{1}{2}$ apart from each other.


Edit: Below is some code to plot the function:

Fs = 100  # some sampling frequency for discretizing the signal to plot it
T = 10    # the time range
t = np.arange(-T, T, 1/Fs)  # the time samples

# define the basis function
f = lambda t: np.exp(-abs(t))

# plot the basis function and the periodic extension
plt.plot(t, f(t))
plt.plot(t, sum(f(t-2*n) for n in range(-10, 10)))




Edit: Regarding your comment:

$$
\begin{align}
\mathcal{F}\left\{e^{−\lvert t\rvert}\right\}&amp;=X_1(j\omega)=\frac{2}{1+\omega^2}\\
\mathcal{F}\left\{\sum_{n}e^{-\lvert t-2n\rvert}\right\}&amp;=\sum_nX(j\omega)\exp\left(j2n \omega\right)\\
&amp;=X(j\omega)\sum_n\exp(j2n \omega)
\end{align}
$$

Here's the trick: 
$$\sum_n\exp(j2n \omega)=\frac{1}{2}\sum_n\delta\left(\omega-\frac{2\pi}{2}n\right)$$

Have a look at the wikipedia article and also this article for example on why this holds.