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:
The Work-Energy theorem tells us that the change in potential energy $\Delta U$ is equivalent to the work done, so:
In your case:
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, $W$.
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.
In order to get to get it up there, you need to provide work against gravity, so:
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:
$$W=\Delta U=mgH=\frac12 mv_0^2$$