1. Let's say that the 3 people are Paul, Graham and Elliot. Paul loses the furst game, then Graham, then Elliot.

At the end, they have 8, 8, 8.

Working backwards, for Paul and Graham to have 8 points after winning, they must have each had 4 before the third game. 4+4 = 8, so Elliot would have lost 8 points altogether.

Therefore, before the third game, they have 4, 4, 16.

Using similar reasoning for the second game shows that before it they each have 2, 14, 8.

Finally, more similar reasoning shows that at the start they have 13, 7 and 4 points respectively.

2. The first few multiples of 13 are 13, 26, 39, 52, 65, 78, 91, 104, 117.

To decide whether a certain number can be scored, look at its last digit. Find the corresponding number in that list and if your number is equal to or bigger than the list number, then it is possible to be scored.

Therefore, the biggest impossible number is the biggest number in that list, minus 10. Namely, 107.

3. I can't really explain how I got this, except to say that I simplified it down so that the total weights were 2, 4, 5, 6, 7, 8, 9, 10, 12 and 13, and then using trial and error.

I got these: 20, 22, 24, 25 and 28.