This is equivalent to the host never opening the door with the car in the Monty Hall scenario
Your error is that you seem to be deciding the answer is 1/3 first, forcing you to assume whatever makes that so. You literally said that when you said the probability must become 1/3.
I do look at this problem from the beginning with no assumptions. If you want to be pedantic, the probability space comprises the sample space (the set of possible outcomes), an event space (a set of subsets of the sample space with certain properties), and a probability function Pr(*) that maps each event in the event space to a number in [0,1]. The complete sample space is {BBb, BBg, BGb, BGg, GBb, GBg, GGb, GGg}, which I assume is what you want (B,G)^3 to mean. We then need the events {BBb, BBg}, {BGb, BGg}, {GBb, GBg}, and {GGb, GGg} to all have probability 1/4.
Since we make no assumptions about the how the lower-case letter is “generated” other than IT MUST MATCH ONE OF THE UPPER CASE ONES, we get that Pr({BBb} = Pr(GGg} = 1/4 and Pr({BBg}) = Pr({GGb}) = 0. What we need to determine is how we get Pr({BGb})+Pr({BGg}) = Pr({GBb})+Pr({GBg}) = 1/4.
In order to make the answer become 1/3 in Q1 and Q2, as you assert, we must assume that we do know how the lower-case letter is generated. In Q1 and Q2 we must assume it is an answer to “is there a girl/boy.” This makes one half of each pair 1/4, and the other 0. If we make no assumption, then the Principle of Indifference (literally, that we make no assumptions to distinguish functionally equivalent outcomes) says Pr({BGb}) = Pr({BGg} = Pr({GBb}) = Pr({GBg}} = 1/8. This makes one answer:
A1 = Pr({GGg}) / [Pr({BGg}) + Pr({GBg}) + Pr({GGg})] = (1/4) / [1/8 + 1/8 + 1/4] = 1/2
Yes, this is a variation of the Monty Hall Problem. Most "solutions" to it are really just explanations for how it can make sense. The mathematical solution follows the outline I used above. It is based on the probability, if the door you choose has the prize (compare to a mixed-gender family), that Monty will open door X or door Y (i.e., the other two) as determined PRIOR TO it happening. If you assume it is 100% for the door he did open, which is only determined AFTER HAVING SEEN IT, like you want to assume in Q1 and Q2 that only a girl/boy can be mentioned, then the answer is that switching does not matter. It is only if you use the Principle of Indifference – meaning each has a 50% chance – that the answer is that switching wins 2/3 of the time.
Take a look at this problem beginning with no assumptions. We have two kids, and an envelope that contains 'B' or 'G'. Our probability space is (B,G)^3, with each having probability of 1/8.
Now, we add information about the match as conditioning. Conditional on being told that the envelope matches the family, we can exclude the BBG and GGB cases. That brings us down to 6, of which we have BBB, GGG, and (BG,GB)(B,G). With this additional information, the probability of matching genders becomes 1/3. This probability is still 1/3 if we open the envelope to find B or G, since we exclude all three cases where the envelope doesn't match our observation of it.
In my view, this is related to the Monty Hall problem; we have to realize that we're given additional information with the statement/envelope.