Answer
a) $\rm 8 \;cycles$
b) From ${\bf 0.938\;\rm MHz} $ to ${\bf 1.03\;\rm MHz}$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The number of complete cycles in one pulse is given by
$$N={\rm\dfrac{Pulse\;length}{Wave\;length}}=\dfrac{L}{\lambda}$$
where $\lambda=v/f$;
$$N =\dfrac{Lf}{v}$$
Plug the known;
$$N =\dfrac{(12\times 10^{-3})(1\times 10^6)}{(1500)}=\color{red}{\bf 8}\;\rm cycles$$
$$\color{blue}{\bf [b]}$$
To find the range of frequencies, or bandwidth, needed to create this pulse, we need to use the concept of the uncertainty principle for signals.
$$\Delta f \approx \frac{1}{\Delta t}\tag 1$$
The pulse duration, in our case, is given by
$$ \Delta t = 8 T_{\lambda} =\dfrac{8}{f}$$
Plug the known;
$$\Delta t=\dfrac{8}{(1\times 10^6)}=\bf 8\times 10^{-6}\;\rm s$$
Plug into (1),
$$\Delta f \approx \frac{1}{ 8\times 10^{-6}}=\bf 1.25\times 10^{5}\;\rm Hz\tag 2$$
Hence the range of frequencies that is needed to be superimposed to create this pulse is given by
$$ \left[f-\frac{1}{2}\Delta f \right] \leq f\leq\left[ f+\frac{1}{2}\Delta f \right]$$
Plug the known from (2) and from the given;
$$ \left[(1.0\times 10^{6})-\frac{1}{2}(1.25\times 10^{5}) \right] \leq f\leq \left[(1.0\times 10^{6})+\frac{1}{2}(1.25\times 10^{5}) \right]$$
$$ {\bf 0.938\;\rm MHz} \leq f\leq{\bf 1.03\;\rm MHz}$$
