I find these problems to be fun. For both problems, we will label the seats 1-6
1: Let's focus on family A, as there are only certain configurations that are valid, and once their configuration is set, it is easy to figure the counts possible.
There can't be an odd number of seats between them, and their low number seat must be odd, or the other two can't be in pairs. Thus, we have 3+2+1 = 3*(3+1)/2 = 6 possible sets of seats they can occupy.
Once A is specified, each of the three couples can be in either order, and B and C can be in either order. Thus, we have 6 * 2^3 * 2 = 96 possible orders
2: The wording of the problem is tricky. if we are strictly considering the women, then they can be in 4 different sets of seats - 1-3 through 4-6 - and in 3! = 6 orders, so the number of ways the women can sit is 4*3! = 4 * 6 = 24 ways.
Note that the men can sit in 3! ways in the remaining chairs, so the 6 people can arrange themselves in 24*6 = 144 different ways.