The probability that the first lefty is the third or fourth person is the sum of the probability that the first lefty is the third person and the probability that the first lefty is the fourth person. The probability that the first lefty is the third person is (0.12)(0.88)(0.12)(0.88)(0.12)(0.12) = 0.0003 and the probability that the first lefty is the fourth person is (0.88)(0.12)(0.12)(0.88)(0.12)(0.12) = 0.0003 so their sum is 0.0006.
The probability that there are exactly 2 lefties in the group is (6 choose 2) * (0.12)^2 * (0.88)^4 = 0.3247.
The probability that there are at least 4 lefties in the group is P(X>=4) = P(X=4) + P(X=5) + P(X=6) = (6 choose 4) * (0.12)^4 * (0.88)^2 + (6 choose 5) * (0.12)^5 * (0.88)^1 + (6 choose 6) * (0.12)^6 * (0.88)^0 = 0.0008 + 0.0001 + 1e-06 = 9e-04.
The probability that there are no more than 2 lefties in the group is P(X<=2) = P(X=2) + P(X=1) + P(X=0) = (6 choose 2) * (0.12)^2 * (0.88)^4 + (6 choose 1) * (0.12)^1 * (0.88)^5 + (6 choose 0) * (0.12)^0 * (0.88)^6 = 0.3247 + 0.3939 + 0.2073 = 92%.