The PAM signal \$s(t)\$ is a weighted sum of functions \$h(t)\$, where the weights are the samples of the signal \$m(t)\$:
$$s(t)=\sum_km(kT_s)h(t-kT_s)$$
This can be modeled as a multiplication of \$m(t)\$ by a comb of Dirac impulses, convolved with \$h(t)\$:
$$s(t)=\left(m(t)\sum_k\delta(t-kT_s)\right)*h(t)\tag{1}$$
From (1) it follows that the spectrum \$S(f)\$ is given by
$$S(f)=\left(M(f)*f_s\sum_k\delta(f-kf_s)\right)\cdot H(f)=
f_s\sum_kM(f-kf_s)H(f)\tag{2}$$
where I've made use of the fact that convolution in one domain corresponds to multiplication in the other domain, and that a Dirac comb in one domain corresponds to a Dirac comb in the other domain (you can find this in most Fourier transform tables). \$M(f)\$ and \$H(f)\$ are of course the spectra of \$m(t)\$ and \$h(t)\$, respectively. So the spectrum \$S(f)\$ is the sum of shifted spectra \$M(f-kf_s)\$, multiplied by the spectrum \$H(f)\$. In order to sketch \$S(f)\$ you need to know \$M(f)\$ and \$H(f)\$:
$$M(f)=\frac{A_m}{2}[\delta(f-f_m)-\delta(f+f_m)]\\
H(f)=T\frac{\sin(\pi Tf)}{\pi fT}e^{-j\pi Tf}$$
For sketching \$S(f)\$ you simply ignore the phase term \$e^{-j\pi Tf}\$ of \$H(f)\$, so you just need to know that the magnitude \$|H(f)|\$ is the magnitude of a sinc function with \$H(0)=T\$ and with zeros at \$f_k=k/T\$, \$k=\pm 1,\pm 2,\ldots\$ (note that \$T\neq T_s\$!).
For (b) just remove all shifted spectra (that's what the ideal low-pass reconstruction filter does), so from (2) you're left with \$f_sM(f)H(f)\$ in the frequency range \$[0,f_s/2]\$.
For question 2 you just need to show that if \$s_1(t)\$ and \$s_2(t)\$ are the PAM signals corresponding to signals \$m_1(t)\$ and \$m_2(t)\$, respectively, then \$as_1(t)+bs_2(t)\$ is the PAM signal corresponding to the signal \$am_1(t)+bm_2(t)\$ for arbitrary constants \$a\$ and \$b\$. This is also obvious because the generation of the PAM signal only involves multiplication and convolution, so it is a linear process.
