Periodic Signal Fourier Transform
The Fourier transform of $x(t)$ can be easily obtained if you realize that $x(t)$ can be written as the convolution of the pulse $R_T(t)$ and a Dirac comb:

$$x(t)=R_T(t)*\sum_{k=-\infty}^{\infty}\delta(t-2kT)\tag{1}$$

Since convolution corresponds to multiplication in the frequency domain, you get for the Fourier transform of $x(t)$

$$X(f)=\mathcal{F}\{R_T(t)\}\cdot\frac{1}{2T}\sum_{k=-\infty}^{\infty}\delta(f-k/2T)\tag{2}$$

because the Fourier transform of a Dirac comb is a Dirac comb (see e.g. this table). And since

$$f(t)\delta(t)=f(0)\delta(t)$$

Eq. (2) can be written as

$$X(f)=\frac{1}{2T}\sum_{k=-\infty}^{\infty}\hat{R_T}\left(\frac{k}{2T}\right)
\delta(f-k/2T)\tag{3}$$

where $\hat{R_T}(f)$ is the Fourier transform of $R_T(t)$. Eq. (3) shows that the Fourier transform of $x(t)$ is a weighted sum of Dirac impulses, where the weights are equidistant samples of the Fourier transform of the pulse $R_T(t)$.

Since the low pass filter $H(f)$ only passes frequencies lower than $3/4T$, only the contributions for $k\in\{-1,0,1\}$ of the sum in (3) are passed. This gives the final result for $Y(f)$. Note that there are two typos in your formulation of $Y(f)$. It should be

$$Y(f)=\frac12\delta(f)+\frac{1}{\pi}\delta(f-1/2T)+\frac{1}{\pi}\delta(f+1/2T)\tag{4}$$