@shibal, you make a pretty good point. While the first answer is not entirely wrong, it only considers the situation where you have exactly the first three cases as your tails.
The way you've presented it makes sense, since it also takes into consideration the option that the three tails appear in other line-ups.
For your second question, the answer will indeed be probability of 3 + probability of 4 + probability of 5. You've already calculated the probability for 3 (5/16). The probability of 4 will be 5/32 or 1/8, since you have five available positions for that stray heads in there:
T T T T H
T T T H T
T T H T T
T H T T T
H T T T T
Getting all tails will obviously be 1/32. Add them all up to get 1/2. This also makes sense if you look at it this way: the probability you're looking for is 1 - (0, 1, 2 tails) which is 1 - (5, 4, 3 heads) or 1 - (the probability of getting at least three heads). Since the probability of getting heads or tails is the same, the probabilities for at least three heads or tails will be exactly the same.
