Answer
$\textbf{(a)}$ $h=0.378\text{ m}$.
$\textbf{(b)}$ It would decrease.
Work Step by Step
$\textbf{(a)}$ The horizontal component of the speed is constant so we can calculate the time required for traveling $l=0.500\text{ m}$ horizontally:
$$l=v_h t\Rightarrow t=\frac{l}{v_h}.$$
Now, since there is no initial vertical component of the velocity, we have for the length of the descend:
$$h=\frac{1}{2}gt^2=\frac{gl^2}{2v_h^2}=\frac{9.81\text{ m/s}^2\cdot 0.500^2(\text{m})^2}{2\cdot1.80^2(\text{m/s})^2}=0.378\text{ m}.$$
$\textbf{(b)}$ If the initial speed is increased, then the sparrow would travel the given horizontal distance for less time $t'$. Since the vertical distance flown is proportional to the time squared it follows that the distance of fall would decrease.
