Since the product of the ages is $72$, the ages must be one of the following combinations:
$$\begin{align*}
&2,2,18\\
&2,3,12\\
&2,4,9\\
&2,6,6\\
&3,3,8\\
&3,4,6\\
\end{align*}$$
The sums are $22,17,15,14,14$, and $13$ respectively. Since the sum and product of the ages didn’t give Jack enough information, the sum must have been $14$, the only possibility that admits more than one solution. The ages must therefore have been either $2,6$, and $6$ or $3,3$, and $8$. If they were $2,6$, and $6$, the two oldest would have been twins, and Bill (probably) wouldn’t have referred to his eldest child: it’s true that one of the six-year olds would technically have been his eldest child, but he’d probably have thought of them as being the same age. Jack inferred that the eldest child wasn’t a twin and concluded that Bill’s children were aged $3,3$, and $8$ years.
Added: Oops! As noted in the comments, $1$ is a possible age. That adds the sets $$\{1,1,72\},\{1,2,36\},\{1,3,24\},\{1,4,18\},\{1,6,12\},\{1,8,9\}$$ to the collection, with sums $74,39,28,23,19$, and $18$; fortunately, these add no further ambiguities.
Gray must have told Frank 6 or greater. (Or else Frank could conclude that the numbers are either $(2,2)$ or $(2,3)$ for 4, and 5 respectively.
Gray must have told Joe a number with 3 or more proper divisors. (Or else Joe can conclude by just finding out the two divisors.)
Suppose the true numbers are $(N,M)$.
Frank is left guessing at $$(2, N+M-2), (3, N+M-3), (4, N+M-4), ...$$
He knows Joe is told one of
$$2(N+M-2), 3(N+M-3), ...$$
But given Joe's answer he can eliminate any of the choices which have too few factors. This must eliminate all but one choice. Then we cannot have had both of these guesses for Frank: $(4, N+M-4)$ and $(6, N+M-6)$, because neither of these can be eliminated by Frank with Joe's information. (Their products both have more than three factors due to the 4 and the 6).
Then $N+M-6 \lt 2 \implies N+M \lt 8$
We can see that $N+M = 6$ does not fit the story:
Then Frank guesses at $(2,4), (3,3)$
Through Joe, he can eliminate both of these options.
Try $N+M = 7$
Then Frank guesses at $(2,5), (3,4)$
After hearing Joe's information, he eliminates $(2,5)$ and knows $(3,4)$. He now knows Gray's numbers. Everyone else in the story then reads this web page and also knows the numbers. So this is the only case consistent with the story.
Best Answer
Well, first off, let's list all the possible combination of ages (and their sum):
$1,1,36; 38$
$1,2,18; 21$
$1,3,12; 16$
$1,4,9; 14$
$1,6,6; 13$
$2,2,9; 13$
$2,3,6; 11$
$3,3,4; 10$
I'm not sure what to make of the building one, but note the specific wording in the third clue: "older". The only reason you would say "older" when referring to THREE people (you would typically use "oldest") means that two of them must be twins. So, you now have three possibilities left:
$1,1,36; 38$
$2,2,9; 13$
$3,3,4; 10$
I don't know how to use the building clue to pare the choices down to one.
That help?
EDIT: Apparently, "older" should be "oldest". In that case, the solution could be any of them but one. In addition, the missing piece is that if the person solving the puzzle knows the number of windows in the building but still cant figure it out, then the two possibilities are:
$1,6,6; 13$
$2,2,9; 13$
At this point, the remark about "oldest" rules out the first one and leaves only $2,2,9$ as the correct answer.