At the top of the hill, gravity ($mg$) acts downwards, and the normal force from the track ($F_N$) acts upwards. The net force provides the downward centripetal force required to follow the circular path. Let the downward direction be positive. Applying Newton's second law:

$$mg - F_N = m a_c = m \frac{v^2}{R}$$

Solving for the normal force:

$$F_N = mg - m \frac{v^2}{R} = m\left(g - \frac{v^2}{R}\right)$$

The critical speed $v_{crit}$ at which the car loses contact with the track occurs when $F_N = 0$. This happens when $g = v_{crit}^2/R$, so $v_{crit} = \sqrt{gR}$.

For Q1, the speed is $v_1 < \sqrt{gR}$. Since $v_1^2 < gR$, the term $(g - v_1^2/R)$ is positive.

$$F_N = m\left(g - \frac{v_1^2}{R}\right)$$

The positive result indicates the normal force is in the assumed upward direction.

For Q2, the speed is $v_2 > \sqrt{gR}$. Since $v_2^2 > gR$, the term $(g - v_2^2/R)$ is negative. A negative normal force is physically impossible, as the track cannot pull the car down. This means the required centripetal force ($mv_2^2/R$) is greater than the available gravitational force ($mg$). The car loses contact with the track, and the normal force becomes zero.

$$F_N = 0$$