>>105797828(1) as there are only 6 terms, and the sequence is (i, j)-separable and the group requires 4 terms (note that m = 1, so we have only one group, and the group must be an arithmetic sequence) the only possible groups are: [1, 2, 3, 4] and [3, 4, 5, 6]
answer: (1, 2), (5, 6)
(2) by exhaustion: if m >= 3, then 4m+2 >= 14. the sequence is (2,13)-separable, therefore the sequence has (at least) the integers {1..14}\{2,13} (so at minimum 12 integers). As we need each group to be an arithmetic sequence, we can divide them into the following 3 (m) groups of 4:
[1,4,7,10],[3,6,9,12],[5,8,11,14]. This suffices, as for all other integers r > 4m+2 in the sequence we can group them as [r, r+1, r+2, r+3] since there will always be groups of 4 integers following 14 if m > 4 as the sequence length is 4m+2.
(3) [unfinished, because I have to get back to work, but I think it can be proven by induction] for m = 1, P1 = [1, 2, 3, 4, 5, 6] this sequence is only (i,j) separable if it i=1, j=2 or i=5, j=6 - see (1). The probability of picking (1,2),(2,1),(5,6),(6,5) { which are (i,j)-separable sequences } is (2*P(2,2))/(6*5) = 4/30, which is (already) greater than 1/8.
for any m, P(m) should (intuitively) be greater P(m-1) That is, the probability only grows as the size of the sequence increases. Combined with P1 that would prove that the probability can never be <= 1/8. But I'll have to think on how to explicitly demonstrate this.