Average Power Spectral Density of PAM signals
Question 1:

If we have a PAM sequence like this:

$a[k]=[0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0]$,

then the PAM signal (Waveform) must hold frequency components which cover these fast transitions from 0 to 1 to 0 to 1. The oscillation speed (frequency) of the transitions are basically given by the time $T$ between the PAM symbols in the waveform.

If $G(f)$ has lower bandwidth than $1/T$, these fast transitions are not possible. All the frequency components which could oscillate at that speed are filtered out by $G(f)$.

Here a plot of the above Waveform, its PAM sequence $a[k]$ and the single pulses (when $a[k]=1$) which summed up create the waveform.



And another plot with the above Waveform and its highest frequency component. If the dashed cosine would oscillate any slower, it could not represent the Waveform anymore.



Question 2:

The minimum bandwidth of $G(f)$ is given by $1/T$, but it can be higher and almost always is (see root raised cosine filter and roll-off factor, which let's you control the bandwidth of $G(f)$). BUT, the PSD of a PAM sequence is only defined within $1/T$, and due to the Nyquist–Shannon sampling theorem it is repeated thereafter (aliasing).

Why is this important?

Look at the following plot, it shows the PSD of a PAM sequence (limited by $1/T$), the PSD of a PAM signal and the filter $|G(f)|^2$.



$G(f)$ filters the PAM sequence and the result is the PAM signal, but what happens outside $1/T$? $G(f)$ filters the aliased part of the PAM sequence.

Note: Not sure if the y-axis of the PSD is correct, but we are only interested in the bandwidth.

Question 3:

You are correct!

But there are some tricks (which e.g. are exploited by the (root) raised cosine filter), see my first plot and the single pulses. They obviously overlap in time, but at each symbol a[k] only one pulse is responsibly for all the amplitude of the waveform. All the other pulses related to other symbols are zero. See the Nyquist ISI criterion.