Robert,
For a) consider the probability of failure  you throw a non6 twice  it's (5/6)^2 or 25/36; thus the probability of success is 11/36.
For b) it might be easiest to see if you made a chart of all 36 possibilities from the two dice when thrown so as to count outcomes.
(1,1) ... (1,6)
(2,1) ... (2,6)
.
.
.
(6,1) ... (6,6)
You succeed if:
you throw (1,6) or (6,1) with the two dice  probability of this is 2/36;
you don't throw (1,6) or (6,1) but do throw a 1 or a 6 and then on the second throw, with the remaining die, you get the other  probability of this is (18/36)(1/6) = 3/36
you do neither of the above and then roll (1,6) or (6,1) on the second throw  probability of this is (16/36)(2/36);
in total p = 2/36 + 3/36 + 32/1296 = 212/1296 only a little more than 1/2 the chances if compared to a) which is 396/1296.
Penny
Cheers,
Penny
