1. When I sum five numbers in every possible pair combination, I get the values: 0,1,2,4,7,8,9,10,11,12.  What are the original 5 numbers?

The sum of the pairwise sums is 64, and this counts each of the original five numbers four times, so the sum of the original five numbers is 16.

The sum of the largest two of the original five numbers is 12, and the sum of the smallest two is 0, so the middle number is 4.

The sum of the largest and middle is the second largest sum, 11, so the largest must be 7, and the second largest is 5.

In order for the second-smallest sum to be 1, one of the numbers has to be -3, so the numbers are -3, 3, 4, 5, 7

2. When I sum a different set of five numbers in every possible group of 3, I get the values: 0,3,4,8,9,10,11,12,14,19.  What are the original 5 numbers?

The sum of the triples is 90, and this counts each of the original five numbers six times, so the sum of the original five numbers is 15.

The largest triple sum is 19, so the two not included sum to -4. The next-largest triple is 14, so the pair not included sum to 1, etc.

So the pairwise sums are -4, 1, 3, 4, 5, 6, 7, 11, 12, 15

The sum of the largest two of the original five is 15, and the sum of the smallest two is -4, so the middle number is 4.

The sum of the largest and middle is 12, so the largest must be 8, and the second largest must be 7.

The sum of the smallest and middle is 1, so the smallest must be -3, so the numbers are -3, -1, 4, 7, 8

3. Is it possible to find a set of 5 numbers in either case above which results in the sums 1-10? 
Find an example or prove it impossible.

No, because the sum of 1-10 must be four times the sum of the five numbers, but the sum of 1-10 is 55, which is odd, and so it is not a multiple of four (as would be required for sums of pairs) or a multiple of six (as would be required for sums of triples). 
Moreover, any sequence of ten consecutive integers sums to an odd number, so no such sequence is possible.

4. If the above problem is not possible, what is the longest series of sequential sums you can find? For example, problems 1 and 2 have six and five sequential sums, (7-12) and (8-12) respectively.

Looking at problem 1, I see that the long sequence is made by taking all the pairs from 3,4,5 and making 7,8,9, then taking pairs of 3,4,5 with 7 to make 10,11,12.

It occurred to me that I could build on this by adding 10 to the end of the sequence, for 3,4,5,7,10 -- this makes pairwise sums 7,8,9,10,11,12,13,14,15,17 -- a sequence of 9. 
Reducing each of the numbers by 2 lowers the resulting sums by 4, so the five numbers 1,2,3,5,8 (F₁ through F₅) give pairwise sums of 3,4,5,6,7,8,9,10,11,13, and the five numbers 0,1,2,4,7 give pairwise sums of 1,2,3,4,5,6,7,8,9,11.

Sums of 3-element subsets of -1,0,1,3,6 are 0,2,3,4,5,6,7,8,9,10. 
Sums of 3-element subsets of -2,1,3,4,5 are 2,3,4,5,6,7,8,9,10,12.

Since no sequence of 10 is possible (using the logic in the answer to problem 3) I conclude this is the best that can be done.