If you move your eyes straight to the bottom you will see that I am not comfortable with what I have done!
So you should probably wait for someone else.
For part A
You would need to know the angle incline of the slope.
Because you need to have the vertical component of the initial velocity
and that would be 96sin(angle) 
I suppose that you would use 2.5 degrees
$$V_y=96sin(2.5^0) \approx 4.187461187 km/hour$$
$${\frac{{\mathtt{4.187\: \!461\: \!187}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{60}}\right)}} = {\mathtt{1.163\: \!183\: \!663\: \!055\: \!555\: \!6}}$$
approx vertical velocity is 1.1632 m/s Initially u=1.1632
a = -9.8m/s^2
Find height (s) when velocity = 0 v=0
Number 4 has u, a, v, and s so use that one
$$\\v^2=u^2+2as\\\\
0=1.16318^2+2*-9.8*s$$
$${\mathtt{0}} = {{\mathtt{1.163\: \!18}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{s}}\right) \Rightarrow {\mathtt{s}} = {\frac{{\mathtt{114\,603\,045}}}{{\mathtt{1\,660\,192\,226}}}} \Rightarrow {\mathtt{s}} = {\mathtt{0.069\: \!029\: \!985\: \!326\: \!530\: \!5}}$$
This is the vertical height that it will coast in metres.
Draw the triangle and you can work out the distance it will coast up the hill
$${\frac{{\mathtt{0.069}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2.5}}^\circ\right)}}} = {\mathtt{1.581\: \!865\: \!408\: \!209\: \!866\: \!1}}$$
1.58 metres - That does not look right
HELP - What did I do wrong ??
Ha ha, I get it!
