What is the change in net internal energy? Suppose a person is hanging on a tree and he falls down to the ground. I consider the person, and the tree as a system and the Earth as surrounding. Applying the energy principle,
$$\Delta K_{trans} + \Delta K_{rel} + \Delta E_{int} = W_{net, ext}$$
What is the work done by external force? Should it be $Mgh$, where $h$ is the height the person is above the ground?
 A: If you are considering the person (and the tree) as the system then as the person falls the external force acting on the person is the gravitational attraction of the Earth on the person $mg$.
If the person (centre of mass) falls a distance $h$ then the work done by the external force on the person is $mgh$ and so the answer to you question is "Yes".
A: Exactly, as this is the only external force acting on the person ( unless you wish to add drag force).
Drag force would be rather negligible though for a person dropping from a tree.
A: No work is done in this situation (assuming no air resistance or drag force).  If a person falls from a ground, he does not lose energy.  Rather, his potential energy (gained from a height above the ground) is converted into kinetic energy (gained from having a speed).  The amount of energy at of the person at the top of the tree will be exactly equal to the amount of energy at the bottom of the tree - this is the Conservation of Energy Law.
You can test it out for yourself if you like.  Assume you have a 1kg person hanging 1m above the ground on a planet with a gravitational force of 1m/s^2 (just to make this simpler).  Let's figure out the initial potential energy above the ground and the final Kinetic energy when the person hits the ground:
Initial:
The formula for potential energy is mgh.  So we plug that in:
mgh = (1kg)(1m/s^2)(1m) = 1 Joule
So the total initial energy is 1 Joule.  This is the value we should also get for final energy.
Final energy:
The formula for kinetic energy is 0.5mv^2.  First we need to figure out the final velocity of the person when he hits the ground:
vf^2 = vi^2 + 2ad
vi must be zero since the person starts at rest when hanging from the tree:
vf^2 = 0 + 2(1m/s^2)(1m)
vf^2 = 2m^2/s^2
vf = 1.414213562 m/s
Now let's plug vf into our formula for Kinetic Energy:
0.5mv^2 = 0.5(1kg)(1.414213562 m/s)^2 = (0.5kg)(2m^2/s^2) = 1 Joule
This proves the initial potential energy and final kinetic energy are exactly equal.
A: You have to consider only one type of energy before calculating the work done. Work done is change in the energy. In this case, You need to calculate the difference in initial and final 'kinetic energy' or initial and final 'potential energy' to get the work done(Ignoring the drag force). 
Please take a look into the below link. 
http://physics.bu.edu/~duffy/py105/Energy.html
A: Here, you consider the tree and the person as the system. Now, what are the forces acting on this system?
The only important force is the gravitational force of attraction exerted by the earth.This force acts on both the bodies. It just manages to have different effects on the bodies in the system. It causes the tree to stay stuck to the ground(, the root system does it but it would be the same if there were no roots, the tree might not have stood upright ,though)while it causes the man to fall down. But only one of these effects causes displacement of a body
The magnitude of this force=mg
The displacement caused=h
Therefore the work done by this force on the system is mgh.
