Answer
$\textbf{(a)}$
$$\vec{g}=(-9.81\text{ m/s}^2)\hat{\bf y}.$$
$\textbf{(b)}$
$$v_0=1.13\text{ m/s}.$$
$\textbf{(c)}$
$$v_0'=3.21\text{ m/s}.$$
Work Step by Step
$\textbf{(a)}$ The acceleration is equal to the gravitational acceleration on the whole road, and thus just before the landing, as well. So it is of magnitude $g=9.81\text{ m/s}^2$ and pointing vertically downwards.
$\textbf{(b)}$
Since the ball rolled off the table, its initial velocity is horizontal. The final vertical component of the velocity is thus:
$$v_v=\sqrt{2gh}$$
The total landing speed is given by
$$v^2=v_h^2+v_v^2=v_0^2+2gh,$$
where we used the fact that $v_h$ is equal to the initial speed. Now we get
$$v_0^2=v^2-2gh$$
and finally
$$v_0=\sqrt{v^2-2gh}=\sqrt{4.0^2-2\cdot9.81\cdot0.75}\text{ m/s}=1.13\text{ m/s}.$$
$\textbf{(c)}$
We will shorten our calculation by using the fact that no matter what is the initial ('horizontal') speed, if the height remains unchanged, the vertical component of the landing velocity will remain unchanged as well.
So, we simply have, by analogy with the last formula
$$v_0'=\sqrt{v'^2-2gh}=\sqrt{5.0^2-2\cdot9.81\cdot0.75}\text{ m/s}=3.21\text{ m/s}.$$
