Taken your interpretation that probability of a team winning, losing and drawing is 1/3.
a) There are 3 games, each game has 3 endings, therefore 3^3 = 27 as the base.
A scores 3, could be only with 1 or with {0,1,2}
only draw is one case, 1+1+1
{0,1,2} has permutations 3! = 6
So in total 7 cases A scores 3 points, probability is 7/27
b) let (x,y) denote points of A then B, following is probability.
(0,0) : 0
(1,1) : 0
(2,2) : 0
(3,3) : 7/27 by above
(4,4) : 0
(5,5) : 0
(6,6) : 0
sum is 7/27
Maybe write down the probability space ($\Omega$,$F$,$P$), where $\Omega$ is the sample space, a family $F$ of subsets of $\Omega$, and a probability measure $P$. For a single event A, if the probability of events is uniform (each event happens with same probability), then use cardinality of A over cardinality of $\Omega$ to calculate the probability.
In problem a), $\Omega$ = {(a,b,c)}|a,b,c $\in${0,1,2}}, $F = 2^\Omega$,A ={(1,1,1),(0,1,2),(0,2,1),(1,0,2),(1,2,0),(2,0,1),(2,1,0)}
For a discrete random vector of more than 1 component, use joint distribution table in general, problem b) is a special case where some thoughts is enough.
