Player 1 e.g. wins with: all odds in any order, then 10 14 4 12 2 8 6
Since each stack must total 24 = even, after playing the odds 1 stack must still be empty (since there are less than 10 odds).
An empty stack cannot be filled with only evens, since 24 cannot be made with a legal combination of evens (if ordered as above)
I determined the sequence of evens as follows: 1/ Write down all (8) possible combos 2/ The only pair (14,10) must be adjacent 3/ 14 can have only one other neighbor so 8,2 or 4,6 must be adjacent (or both); same for 12,2 and 8,6 - a guess: using 2,8,6 seems a good choice since it helps for both requirements.
4/ Trying to add the now most problematic 12 to 14,10 and 2,8,6 yielded my solution after some fiddling. - note that there are many? more possibilities, e.g. 4,12,2,8,6,14,10
Addition: Bonus question
The odds are in player 1s favor
I don't see a guaranteed win, but again player one can make sure the evens do not make the required total of 30: 4 14 10 12 8 6. (note: 2 is not needed to prevent 30, it can be 1st or 7th) If the odds are put behind those evens in random order, player 1 has 47/70 chance of winning according to my calculations.
Note that 2 odds must be assigned to each stack, as soon as the first is assigned, the other is fixed, and may be the next coming up, giving player 1 the win.
summary of calculation that sums up to 47/70:
Looking at the pairing of the first 4 odds coming up:
pairs chance chance of success for player 1
1234 - 8/35 - 1/4 (fifth odd is from pair 4)
1212 - 4/35 - 2/3 (only 3434 is a win for player 2)
1231 - 4/35 - 1/2 (pair 4 must be adjacent, 44XX X44X and XX44 are success)
1232 - 4/35 - 1/2 (same)
1213 - 4/35 - 7/12 (also 3424 is successful)
other- 11/35 - 1 (already an adjacent pair within the first 4 odds)