Given a signal and its Fourier transform, find FS coefficient of the shifted sum of the signal
Figure out that $x_2(t)$ is basically, sum of shifted copies of $x_1(t)$ which can be written as follows :
$$x_2(t) = x_1(t) * \sum^{\infty}_{k=-\infty} \delta(t-6k)$$, where, $*$ represents convolution operation. therefore, the Fourier Representation of $x_2(t)$ will be product of the fourier representation of $x_1(t)$ and the pulse train $\sum^{\infty}_{k=-\infty} \delta(t-6k)$.

Fourier Transform of a pulse train $\sum^{\infty}_{k=-\infty} \delta(t-6k)$ is given by the following :

$$\mathscr F \Big\{ \sum^{\infty}_{k=-\infty} \delta(t-6k) \Big\} = \frac{1}{6} \sum^{\infty}_{k=-\infty} \delta(f-\tfrac{k}{6})$$

Which basically means that Fourier Representation of $x_2(t)$ becomes :
$$X_2(f) = X_1(f) \cdot \frac{1}{6} \sum^{\infty}_{k=-\infty} \delta(f-\tfrac{k}{6})$$

So, $f = \frac{1}{6}$ is the fundamental frequency, since you have periodized $x_1(t)$ by $T = 6$. And, therefore, at all multiples of $\frac{1}{6}$, you will get a Fourier coeff, which will be equal to $\frac{1}{6} X_1(\frac{k}{6})$.

What this means is, $X_2(f)$ is nothing but sampled version of $X_1(f)$ at $f = \frac{k}{6}$ and hence, $X_2(f)$ is discrete. Which is evident from the fact that $x_2(t)$ was periodized in the first place.