Due to the symmetry of projectile motion, the time spent ascending through any height segment is equal to the time spent descending through it. We can find the total time by calculating the descent time and doubling it.

Consider the object falling from rest from its peak height $H$. The time $t$ to fall a vertical distance $y$ is given by the kinematic equation $y = \frac{1}{2}gt^2$, which yields $t(y) = \sqrt{2y/g}$.

[Q1] The time to fall through the top segment of height $h$ is simply the time to fall a distance $y=h$.

$$t_{fall, top} = t(h) = \sqrt{\frac{2h}{g}}$$

The total time is twice the fall time.

$$T_{top} = 2t_{fall, top}$$

[Q2] The time spent falling through the bottom segment is the total time to fall the full height $H$ minus the time to fall the distance from the peak to the top of the segment, which is $H-h$.

$$t_{fall, bottom} = t(H) - t(H-h) = \sqrt{\frac{2H}{g}} - \sqrt{\frac{2(H-h)}{g}}$$

The total time is twice this fall time.

$$T_{bottom} = 2t_{fall, bottom}$$