Suppose there is a family with two children. The family tells me that at least one child is a boy, and wants me to guess the gender of the other child. Let's say there are two things I can do: I can either guess right away, or I can ask whether the boy is the first or the second child, and then I can guess accordingly. If both children are boys, then they answer randomly with first or second.
If in both cases my guess is that the other child is a girl, here are the possibilities for each case:
Case 1: Because I do not know which child is a boy, the possibilities are BB, BG and GB. Since they are all equally probable, the probability that I got right is $\frac 2 3$.
Case 2: In this case I asked whether the boy is the first or the second child. If they give the answer 'first', then the possibilities are BB and BG, so the probability that I got right is $\frac 1 2$. This would be equivalent to the case in which I knew beforehand that the first child is a boy and guessed accordingly. If they give the answer 'second', then the probability that I got right is analogously $\frac 1 2$.
So, in case 1 the probability that I got right is $\frac 2 3$, and in case 2 the probability is $\frac 1 2$, although in both cases I guessed 'girl' no matter what the answer I got in case 2 is. Just the mere fact that I asked that question and got an answer to it reduced my possibility of getting the answer right. How is this possible? There must be something I am missing or miscalculating.
Here is your case 1:
$$\begin{array}{|c|c|c|} \text{Older} & \text{Younger} & \text{Statement 1} & \text{Probability} \\ \hline \text{Girl} & \text{Girl} & \text{No boys} & \frac14\\ \hline \text{Boy} & \text{Girl} & \text{At least 1 boy} & \frac14\\ \hline \text{Girl} & \text{Boy} & \text{At least 1 boy} & \frac14\\ \hline \text{Boy} & \text{Boy} & \text{At least 1 boy} & \frac14\\ \hline \end{array}$$
so the conditional probability given the statement "At least $1$ boy" of both being boys is $\dfrac{\frac14}{\frac14+\frac14+\frac14}=\dfrac13$.
Here is your case 2:
$$\begin{array}{|c|c|c|} \text{Older} & \text{Younger} & \text{Statement 1} & \text{Statement 2} & \text{Probability} \\ \hline \text{Girl} & \text{Girl} & \text{No boys} & \text{Neither is a boy} & \frac14\\ \hline \text{Boy} & \text{Girl} & \text{At least 1 boy} & \text{Older is a boy} & \frac14\\ \hline \text{Girl} & \text{Boy} & \text{At least 1 boy} & \text{Younger is a boy} & \frac14\\ \hline \text{Boy} & \text{Boy} & \text{At least 1 boy} & \text{Older is a boy} & \frac18\\ \hline \text{Boy} & \text{Boy} & \text{At least 1 boy} & \text{Younger is a boy} & \frac18\\ \hline \end{array}$$
so the conditional probability given the statement "Older is a boy" of both being boys is $\dfrac{\frac18}{\frac18+\frac14}=\dfrac13$.
Similarly for the statement "Younger is a boy".
Note that in case 2, the calculation would have been different if the second question had been "Is the older child a boy?" so with two boys they always said the older was a boy, i.e. if the fourth row had probability $\frac14$ and the fifth row had probability $0$. The conditional probability would have become $\frac{\frac14}{\frac14+\frac14}=\frac12$.