Answer
See the detailed answer below.
Work Step by Step
a) We need to draw the force diagram of the player as he is in the action of jumping before his legs leave the floor. See the first figure below. You can see that the normal force exerted by the ground on the player is equal to the force exerted by the player on the ground since they are an action-reaction pair.
b) Just before the feet of the player leave the ground, the normal force exerted on him by the ground is greater than his own weight, as we see in figure b below. So, yes, there is a net force exerted on him upward that allows him to jump into the air.
c) We can calculate his acceleration since we know he traveled upward of 60 cm while touching the ground before lifting off and since we know that he traveled 80 cm in the air after lifting off.
The lifting-off speed is given by the kinematic
$$\overbrace{v_{2y}^2}^{0}=v_{1y}^2+2a_y\Delta y_2$$
The final speed is zero and his acceleration in the air is the free-fall acceleration.
$$0=v_{1y}^2-2g\Delta y_2$$
Thus,
$$v_{1y}=\sqrt{2g\Delta y_2}\tag 1$$
$$v_{1y}=\sqrt{2\times9 .8\times 0.8} =\bf 3.96\;\rm m/s$$
And his acceleration while jumping is given by
$$v_{1y}^2=\overbrace{v_{iy}^2}^{0}+2a_{y1}\Delta y_1$$
Thus,
$$a_{y1}=\dfrac{v_{1y}^2}{2\Delta y_1}$$
Plugging from (1);
$$a_{y1}=\dfrac{\color{red}{\bf\not}2g\Delta y_2}{\color{red}{\bf\not}2\Delta y_1}=\dfrac{ g\Delta y_2}{ \Delta y_1}$$
$$a_{y1} =\dfrac{ 9.8\times 0.8 }{ 0.6}=\color{red}{\bf 13.1}\;\rm m/s^2$$
e)
The scale reads the normal force exerted on the person.
So, the reading of the scale before he jumps is his own weight.
$$\sum F_y=F_{n1}-mg=ma_{y}=m(0)=0$$
$$F_{n1}=mg=100\times 9.8=\color{red}{\bf 980}\;\rm N$$
The reading, while he jumps, is
$$\sum F_y=F_{n2}-mg=ma_{y1}$$
$$F_{n2}=ma_{y1}+mg$$
$$F_{n2}=(100\times 9.8)+(100\times 13.1)=\color{red}{\bf 2290}\;\rm N$$
The reading while he is in the air, is zero since he no longer touches the scale.
$$F_{n3} =\color{red}{\bf 0}\;\rm N$$
