Hemophilia is an X-linked recessive trait in humans. Huntington’s Disease is inherited with an autosomal dominant allele.
Let 'a' be the recessive allele for Hemophilia, 'A' the corresponding dominant allele. It is X-chromosome linked recessive gene, thus males are haploid for this gene.
Let 'B' be the dominant allele for Huntington's, and 'b' the corresponding recessive allele. It is autosomal dominant.
(Note: using different capitalisations of the same letter for the two diseases seems to have muddled your logic.)
a. Mr. Y is unaffected by either condition. He marries Ms. X, who is unaffected by hemophilia but shows signs of Huntington’s Disease. Ms. X’s father has hemophilia but is unaffected by Huntington’s Disease. What is the probability that the first child born to Ms. X and Mr. Y will be a son with hemophilia and their second child will be a daughter with Huntington’s Disease and their third child will be a son who is unaffected by either condition?
First a little deductive work.
Mister Y is 'A bb' - neither a dominant Huntington's gene, nor a recessive haemophilia gene.
Father X is genotype 'a bb' - no dominant Huntington's gene, but a recessive haemophilia on the X-chromosome.
Miz X is thus 'Aa Bb' - since she must receive the recessives from her father, and the dominants from her mother, thus showing Huntington's and not being a haemophiliac.
So the crossings are: $\rm 'A'\times'Aa' = \{\underbrace{'AA','Aa'}_{\text{female}},\underbrace{'A','a'}_{\text{male}}\}$ and $\rm 'bb'\times 'Bb' = \{'Bb', 'bb'\}$.
That is their children have a $50\%$ probability of showing Huntington's, their daughters will not be haemophiliac but have a $50\%$ probability of being carriers, but their sons have a $50\%$ probability of being haemophiliacs.
Notes:
You assumed that Father X and Mr. Y both had to be either homozygous recessive or heterozygous yet not showing with equal probability, but this does not follow.
Because we are given no measure of how likely someone with the gene is to develop the disease (or not), nor how prevalent the disease is in the general population, we need to assume Complete Dominance. This is that that 'shows no sign of Huntington's', or 'unaffected' means 'does not have a dominant gene for it'.
Otherwise we can't make any meaningful assessment on the probabilities of Mr. Y or Father X having a dominant gene for the trait, let alone Mother X.
In the two-frog scenario, the event "One croak was heard" is not the same as the event "at least one frog is male".
There are eight possibilities: the "left" frog can be female or male, the "right" frog can be female or male, and exactly one croak was emitted or not. Assuming males croak with probability $p$ while you are in their presence, and everything is independent, the following table enumerates the eight possibilities and their probabilities:
$$
\begin{array}{c|l|c|c|c|c}
\text{Outcome} & \text{Probability} &\text{One croak?}&\text{F present?}&\text{At least one M?}\\
\hline
FF0 & \frac14 & & Y&\\
FF1 &0&Y&Y&\\
FM0 &\frac14(1-p) &&Y&Y\\
FM1&\frac14p&Y&Y&Y\\
MF0&\frac14(1-p)&&Y&Y\\
MF1&\frac14p&Y&Y&Y\\
MM0&\frac14(p^2+(1-p)^2)&&&Y\\
MM1&\frac14\cdot 2p(1-p)&Y&&Y\\
\end{array}
$$
Using the above table, the probability that you survive given you heard a croak is
$$
P(\text{F present}\mid\text{one croak})=\frac{P(FF1,FM1,MF1)}{P(FF1,FM1,MF1,MM1)}=\frac{0+\frac14p+\frac14p}{0+\frac14p+\frac14p+\frac142p(1-p)}=\frac1{2-p}.
$$
This makes intuitive sense, because if males croak all the time ($p=1$), then for sure the other frog is female; if males croak but very rarely, then it's a coin toss whether the other frog is female. Note that the above probability is never smaller than $\frac12$, so the two-frog lick is always a better strategy than the one-frog lick.
Using the above table, the probability $P(\text{survive}\mid\text{at least one M})$ is properly calculated as $2/3$, but this result is neither here nor there, because you didn't observe that event.
As for your other question, we can modify the outcome space to specify whether the left frog croaked or the right one. Using a similar enumeration to the above, the prob that you survive given you heard a croak from only the left frog is:
$$P(\text{F present}\mid\text{only left frog croaked})
={P(M1F0)\over P(M1F0,M1M0)}={\frac12p\cdot\frac121\over\frac12p\cdot\frac121+\frac12p\cdot\frac12(1-p)}=\frac1{2-p},
$$
exactly the same probability as when you didn't know which frog croaked. The probability of survival given that only the right frog croaked is the same. Conclusion: knowledge of which frog croaked does not lower your probability of survival.
Best Answer
The answer that @Lordofdark linked to is very interesting. In addition to the matter of the probability of croaking, there are other factors to consider. For instance, the problem doesn't specify whether the sex of the frogs is independent. For all we know, male frogs might never be found alone, so we should definitely lick the solitary frog. Or perhaps it's mating season and pairs of frogs are always male/female. Perhaps male frogs only croak when they're around females. Or...
Simply put, there is not enough information to go on. A mathematician lost in the jungle will probably die from taking too long to consider the possibilities.