In case of 4 people, the immediate neighbours of the starter each have a starting probability of 1/2 of not being the last.
For the one opposite, it does not matter whether the first toss is heads or tails. After the first toss, he also has a probability of 1/2 of not being the last. So his effective starting probability is also 1x1/2 = 1/2
All have equal probability of getting the coin before the last person.

If he eats apples only if they are available, it is much simpler then
Although a better way could be for the driver to drive an empty truck to the local store and explain the problem. Let the local store send a non-apple-eating driver with the truck who makes 3 trips delivering all the 3000 apples.

Bonanova is right. In the above solution, the number of cases where the last person does not get his seta is much higher than n-1. It is actually 2^(n-2) considering all possibilities where the "loop is closed" with n-2 persons instead of n-1 so that the last person's seat is always available.
The probability will then be as suggested by Bonanova.

Clarification needed on owing the money:
If for example, I choose B and the computer also selects B, what happens?
a) I get $1000 from the envelop and I return those $1000
b) I get nothing but still owe $1000 to my uncle?

I don't see why 90° should be an issue. There are no coins that make 90° -- 10 sides for 360°; angle between any 2 vertices is a multiple of 36° and will never be 90°.
I agree with gavin on the solution.