hello..if it's not pretty late then here's my reply:
(btw, i reckon both of these are GMATPrep quesn)

Q1: p(A,B) is calculated by:
choosing AorB out of 6 workers: 2/6
now, let's say we have either one of them chosen up, then the remaining ways of choosing the other is 1/5.
since both of these are independent events. thus, (2/6)*(1/5) = 1/15

Q2: here only multiplication of 'x+y' & 'x-y' gives me the required form of resulting expression x^2 - (1*y)^2
so similarly, chances of choosing either of them: 2/4 and once one is in the bag, chance of choosing other is: 1/3
therefore, combining them, we get the result (just like prev quesn)
2/4 * 1/3 = 1/6 [ And, i know this one is right for sure ]

All the best, Sandman

1) 1c1.1c1 / 6c2 = 1/15

2) On a smilar note the prob = 1c1.1c1/4c2 = 1/6

Alternate solution:

P(x+y) picked * p(x-y) picked + p(x-y)picked * p(x+y) picked

= 1/4*1/3+1/4*1/3
=2/12
=1/6

I am pretty sure these are the solution but kindly wait for other answers too just to be 100% sure.

All the best for your exam; Be confident and succeed!!!