Work done by gravity on falling object does not seem to equal change in mechanical energy So I have some confusion here, I am sure I knew this at some point.  Let's say an object of 10 kg is dropped from a height of 10 m.  When it reaches the ground, the work done on the object should be the force ($mg$) x distance or 10 kg x 9.8 m/s/s x 10 m.  That gives 980 joules of work done on the object by gravity.
But the object did not gain 980 joules of mechanical energy.  It lost 980 joules of GPE and gained 980 joules of kinetic energy (up to the point of it reaching ground level). 
Using the change in GPE and KE, it looks like no work was done on the object because the loss in GPE equals the gain in KE.
So 
a) am I right that no net work was done on the car by sum of all forces?  
b) is the work done by gravity equal to force time distance, or is it equal to the change in mechanical energy of the object which is zero?   
 A: The confusion here comes from the fact that your choice of system is not clearly defined.
If the system is the earth plus the object, then there is no external force, and therefore no change in total energy.  The potential energy of the system is transfered into kinetic energy.  No external work done, and external work is what adds or removes energy of the system.  
If the system is the object, then gravity does external work on the system, adding energy, increasing its kinetic energy.   Potential energy is not defined for a single object.  There is no potential energy with this choice of system.  Potential energy is always defined for pairs of interacting objects.  With this system, there is work done.
A: 
So a) am I right that no net work was done on the car by sum of all forces? b) is the work done by gravity equal to force time distance, or is it equal to the change in energy of the object which is zero?

You defined work correctly for your example:
$$W=F\Delta h=mgH$$
The Work-Energy theorem tells us that the change in potential energy $\Delta U$ is equivalent to the work done, so:
$$W=\Delta U$$
In your case:
$$\Delta U=mgH$$
So it tallies perfectly!
It's clear you're confusing conservation of energy with work-energy equivalence. Work was done but overall the total energy of the system hasn't changed: only potential energy, $U$, has been converted to work, $W$.

Edit: in answer to OP's last comment.


1) a 10 kg object is slowly raised to a height of 10 m. Its potential energy has increased by 980 J, it is motionless, and so I assume that you did 980 J of work on the object to raise it. 2) You toss the object up, it gains enough kinetic energy to rise to a height of 10 m before it stops rising. It is in the same final state as situation 1, but didn't gravity do -work to decelerate it? You do the same work to accelerate it so that it rises to 10 m and motionless as if you raised it slowly.

Case 1):
In order to get to get it up there, you need to provide work against gravity, so:
$$W=mgH=\Delta U$$
Case 2):
You toss the object up and it just reaches $H$.
To do so, you will have to impart kinetic energy $\Delta K$ to the object, equivalent to $\Delta U=mgH$, so $\Delta K=\Delta U$. During 'flight' this kinetic energy is then converted to potential energy and the object ends up with $K=0$ because $v=0$.
To impart that kinetic energy $\Delta K=\frac12 mv_0^2$ ($v_0$ is the launch velocity), you need to perform work:
$$W=\Delta K=\frac12 mv_0^2$$
And since as $\Delta K=\Delta U$, then:
$$W=\Delta U=mgH=\frac12 mv_0^2$$
A: This is a common point of confusion that boils down to the fact that there are two physically equivalent but conceptually different ways of viewing this situation.
You can either look at gravitational potential energy (GPE) as an "internal" form of energy that your 10 kg object can have or you can look at the gravitational force (Fg) as an external force acting on the object. What you cannot do is look at the situation in both ways at once.
If you choose to see GPE as a form of energy, what is happening in this situation is that GPE is turning into kinetic energy (KE) as the object falls. If you choose to see Fg as an external force, what is happening is that Fg is doing work on the object, which increases its kinetic energy.
Either way, the amount by which the kinetic energy increases when the object reaches the ground is mgh, or 980 Joules in this case.
