You posed two questions:
First, Earth exerts a gravitational force on the car. The car, in turn, exerts a gravitational force on Earth. These are equal and opposite. This can be seen from the fact that the formula for magnitude of the (Newtonian) gravitational force:
$$F_g=\frac{GM_1M_2}{r^2}$$
remains of exactly the same form if the masses are switched. The fact that they are acting in opposite directions is obvious.
Furthermore, one could say that the car exerts a fore upon the ground. This force is just because of the attraction between Earth and the car. Then, the reaction force is the contact force that the ground exerts on the car, keeping it in its place.
About the answer you found, I can say the following things. You probably know that $\vec{F}_{\text{net}}=m\vec{a}$. But we can write $\vec{a}=\frac{d\vec{v}}{dt}$ where $\vec{v}$ is the velocity. Since $m$ is constant, we can say
$$\vec{F}=m\vec{a}=m\frac{d\vec{v}}{dt}=\frac{d(m\vec{v})}{dt}=\frac{d\vec{p}}{dt}$$
So, you see that the force is the time derivative of the momentum $\vec{p}=m\vec{v}$. If the momentum is constant, the time derivative, and therefore the net force, is zero. Therefore, all forces acting on the car must cancel out. Since there are only $2$ forces, they must be equal and opposite in order to cancel out, so we arrive at the conclusion.
