Answer
$\textbf{(a)}$ $d=1.55\text{ m}$;
$\textbf{(b)}$ The drop decreases;
$\textbf{(c)}$ The drop decreases.
Work Step by Step
Let us denote $L=18$ m.
$\textbf{(a)}$ The horizontal distance is covered with the constant speed $v_h=32\text{ m/s}$ so we have for the time of flight:
$$t=\frac{L}{v_h}.$$
The drop is due to gravitational acceleration. Since there is no initial velocity in vertical direction we have for the drop
$$d=\frac{1}{2}gt^2=\frac{gL^2}{2v_h^2}=\frac{9.81\text{ m/s}^2\cdot18^2\text{ m}^2}{2\cdot 32^2(\text{ m/s})^2}=1.55\text{ m}.$$
$\textbf{(b)}$ The drop decreases because with the increase of the speed, the time required for the pitch to be caught is less so the drop is also decreased.
$\textbf{(c)}$ The drop would decrease because the gravity on the Moon is weaker than on the Earth (the gravitational acceleraton is lower) so the drop will be lower.
