Atennis ball is served from the back line of the court such that it leaves the racket 2.4 m above the ground in a horizontal direction at a speed of 22.3 m/s.

This is a projectile motionm with

Iniital height, Yo = 2.4 m
Iniital vertical velocity, Voy = 0
Kind of motion: vertical fall => Y = Yo - Voy * t - [g*t^2] / 2

Initial horizontal position, Xo =0
Initial horizontal velocity, Vox = 22.3 m/s
kind of motion: uniform in the horizontal direction => X = Vox *t

You can find:

a) Height when the ball is passing the net, which is at 11.9 m from the service point.

Time when X = 11.9 m

X = Vox*t => t = X / Vox = 11.9 m / 22.3m/s = 0.5336 s

Replace that time into the vertical equation to find the height, Y:

Y = Yo -  Voy*t - g*(t^2) / 2 = 2.4m - 0 - 9.8 m/s^2 * (0.5336 s)^2 / 2 = 1.0 m

Then, given that height of the net is 0.91 m, the ball will pass 0.09m over the net (9 cm).

b) How far from the net the ball will hit the ground,  given that it is 11.9 m away of the service point.

From the vertical equation you can determine the flight time:

Y = Yo - g(t^2) / 2 => 0 = 2.4 - 4.9 t^2 => 4.9t^2 = 2.4 = > t^2 = 2.4 / 4.9 = 0.49 => t = √0.49 = 0.7 s

Now use that time in the vertical equation to find the X position of the ball when it hits the ground:

X = Vox *t = 22.3m/s * 0.7 s = 15.6 m  

Then, subtract the position of the net to find how far away from it the ball hits the ground: 15.6 - 11.9 =3.7 m.