Answer
$F=2066N$
Work Step by Step
We take the road along the hill to be our horizontal (x) axis, with upward being the +x direction. Since the car's motion is also along this axis, we have
$$\sum W=\sum F_{x}s (1)$$
$\vec{F_N}$ is directed perpendicular with the car's displacement, so it plays no role in the net work.
On the other hand, $\vec{F}$ and $\vec{f}$ both lie on the horizontal and contribute to $\sum \vec{F_x}$. $\vec{F}$ propels the car upward, while $\vec{f}$ opposes it.
Besides $\vec{f}$, the x component of the car's weight $W\sin5$ also opposes the car's movement. Therefore, we have $$\sum F_x=F-f-W\sin5$$
Plug this back to (1): $$(F-f-W\sin5)s=\sum W$$ $$F=\frac{\sum W}{s}+f+W\sin5$$
We have $\sum W=+150kJ=1.5\times10^5J$, $f=524N$ and $s=290m$
The car's weight is $W\sin5=1200\times9.8\sin5=1024.95N$
Therefore, $$F=2066N$$
