What is the probability that three babies are boys, given that at least one is a boy?

In the second problem, please note your set excludes the case where none of the fetus is a baby boy as you know at least one of them is.

So the probability in the second case that all three fetuses are baby boys $ \displaystyle = \frac{0.485^3}{1-0.515^3}$

Let $X\in\{0,1,2,3\}$ be the number of boys. Let $A$ be the event that $X=3$ and $B$ be the event that $X\ge1$. By definition, we have ${\rm P}(A\cap B)={\rm P}(A\mid B){\rm P}(B)$ so $${\rm P}(X=3\mid X\ge1)=\frac{{\rm P}(X=3\cap X\ge1)}{{\rm P}(X\ge1)}=\frac{{\rm P}(X=3)}{{\rm P}(X\ge1)}.$$ You have calculated ${\rm P}(X\ge1)$ in the first part and evaluating ${\rm P}(X=3)$ is straightforward.

now, I am not sure about an "official" sol but here is mine.

now in these cases, I would use sets and a Venn diagram. let's name the persons 1,2,3 now let A = {events|1 has a boy} and similarly for B and C so the Venn diagram would look like-Venn diagram

so now the if at least one of them has a boy then we are in A$\cup$B$\cup$C and so the required probability is n(A$\cap$B$\cap$C)/n(A$\cup$B$\cup$C) = $0.485^3/(1-0.515^3)$