Assume X = rolling a 6 (also / indicates a fraction, ^ indicates exponent)

for a) P(X=1)= (6C1)(1/6)^1(1- 1/6)^6-1
= (6C1)(1/6)(5/6)^5
= (6)(1/6)(3125/7776)
= 3125/7776
Therefore the probability of rolling a 6 in exactly one of the six throws is 3125/7776

for b)
P(X > or = 1)= 1- P(X=0)
=1- ((6C0)(1/6)^0(1-1/6)^6-0)
=1- ((1)(1)(5/6)^6)
=1- ((1)(1)(.334897976))
=1-(.334897976)
= .665102023
Therefore the probability of rolling at least 1 six in six rolls is .665