(b) To have no winner within the first 6 tosses means that there must be 3 tails and 3 heads tossed. So we count how many ways can that happen. The number of ways to choose 3 heads (or tails, doesn't matter) from 6 tosses is

${6\choose 3}=\frac{6\cdot 5\cdot 4}{3\cdot 2} = 20$

so 20 ways winthin the 6 tosses. but the question was "how many ways among the 128 of 7 tosses" so we need to add the 7th toss by multiplying by another 2 possibilities for the outcome of the 7th toss, giving us the answer: among the 128 there are 40 games that are tied and need the 7th toss.

(c) The probability is the number of tied games (after 6 tosses) divided by the total number of possible configurations among 6 tosses:

$P(\mbox{"7th toss needed"}) = \frac{20}{64} = \frac{5}{16}$