Answer
$\textbf{(a)}$ $h=1.41\text{ m}$.
$\textbf{(b)}$ It would remain the same.
Work Step by Step
$\textbf{(a)}$ The horizontal distance is covered with the constant horizontal component of the velocity (which is thus equal to the initial velocity, having the magnitude of $v_h=3.32\text{ m/s}$), so we have
$$l=v_h t\Rightarrow t=\frac{l}{v_h},$$
where $l$ is the horizontal distance, and $t$ is the time of flight.
Now, since there is no initial vertical component of the velocity, we have for the height:
$$h=\frac{1}{2}gt^2=\frac{gl^2}{2v_h^2}=\frac{9.81\text{ m/s}^2\cdot1.78^2\text{ m}^2}{2\cdot3.32^2(\text{m/s})^2}=1.41\text{ m}$$
$\textbf{(b)}$ The time of flight, even though deceivably depends on $l$ and $v_h$, actually does not depend physically on either of them (this dependence will cancel). Namely, since the initial velocity is strictly horizontal, it matters only how high above the water the board is because the vertical distance is covered due to the gravitational acceleration only and reducing the initial speed will have no effect on the time of the flight.
