Answer
(a) $\frac{F_N}{W_{car}}=0.97$
(b) $\frac{F_N}{W_{car}}=0.82$
Work Step by Step
From the picture below, we can see that the normal force $\vec{F}_N$ does not point directly upward but points in a direction perpendicular with the inclined surface. Therefore, the opposite force of $\vec{F}_N$ is no longer $m\vec{g}$, but $m\vec{g}\cos\theta$
Considering only $\vec{F}_N$ and $m\vec{g}\cos\theta$, we see that these two forces do not contribute to the car driving. In fact, they balance each other: $$F_N=mg\cos\theta$$ $$\frac{F_N}{mg}=\frac{F_N}{W_{car}}=\cos\theta$$
(a) $\theta=15^o$
$$\frac{F_N}{W_{car}}=\cos15=0.97$$
(b) $\theta=35^o$
$$\frac{F_N}{W_{car}}=\cos35=0.82$$
