I think it is explained in your OP. We do the same thing with a bit more detail.
Since we have already counted the number of "bad" positions with all the boys together, it remains to count the number of bad positions in which the boys are not all together, but some boy is not next to a girl.
There must be two boys together, and they must be at the left end or the right end ($2$ choices). Take such a choice, say left end. Then the remaining boy can be in any one of $3$ places, namely positions $4$, $5$, or $6$. That gives us $(2)(3)$ ways of choosing the seats the boys will occupy. For each of these $(2)(3)$ choices, there are $3!$ ways of permuting the boys among the chosen seats, and for every such choice there are $3!$ ways of permuting the girls, for a total of $(2)(3)(3!)(3!)$.
Another way: Counting all possible arrangements and subtracting the "bad" ones is often good strategy. But let us count directly.
Our condition will be satisfied if either no two boys are together, or exactly two boys are together and they are not in the end seats.
Count first the arrangements in which no two boys are adjacent. Write down $G\quad G\quad G$. This determines $4$ "gaps" into which we can slip the boys, one boy per gap. There are $4$ "gaps" because we are including the end gaps. There are $\binom{4}{3}$ ways of choosing $3$ of these gaps.
Or else we could slip $2$ boys into one of the two center gaps ($2$ choices), and then slip the remaining boy into one of the $3$ remaining gaps, for a total of $6$ choices.
Thus the places for the boys can be chosen in $10$ ways. For each of these ways, we can arrange the boys is $3!$ ways, and then the girls in $3!$ ways, for a total of $360$.
Keep in mind that seating arrangements at a circular table are invariant with respect to rotation. Thus, once we seat the first person, all permutations are relative to where that person sits.
(a) Let's sit the girls first. One girl sits down. There are $3!$ ways to seat the other girls. Once the girls sit down, there are $4!$ ways to arrange the boys, giving
$$3!4! = 6 \cdot 24 = 144$$
possible seating arrangements in which the boys and girls alternate seats.
Alternate solution: There are $4!$ ways of sitting the girls. However, there are four rotations of the girls that do not change their relative order. Thus, there are
$$\frac{4!}{4} = 3!$$
distinguishable ways of sitting the girls. Once they are seated, there are $4!$ ways of sitting the boys. Thus, there are $4!3! = 144$ ways of sitting the girls in which the boys and girls alternate seats.
(b) We seat the boy and girl in adjacent seats first. Once she sits down, he can sit either to her right or to her left, giving two ways of arranging the boy and girl who must sit in adjacent seats. Relative to her, there are $3!$ ways of seating the remaining girls. Once they are seated, there are $3!$ ways of seating the remaining boys, giving
$$2! \cdot 3! \cdot 3! = 2 \cdot 6 \cdot 6 = 72$$
seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl sit in adjacent seats.
Alternate solution: We showed above that we can seat the girls in $3!$ distinguishable ways. There are $2$ ways we can sit the particular boy next to the particular girl (on her right or her left). Once he has been seated, there are $3!$ ways to seat the remaining boys, which yields
$$3! \cdot 2 \cdot 3! = 6 \cdot 2 \cdot 6 = 72$$
seating arrangements in which a particular boy sits next to a particular girl.
(c) We subtract the number of ways we can sit a particular boy and girl in adjacent seats when the boys and girls sit in alternate seats from the total number of ways they can sit in alternate seats, which yields
$$3!4! - 2!3!3! = 144 - 72 = 72$$
seating arrangements in which the boys and girls sit in alternate seats and a particular boy and girl do not sit in adjacent seats.
Best Answer
I think of it this way. One of the girls is the Queen, and one of the chairs is a throne. The Queen of course sits on the throne. The rest of the girls can be permuted in $3!$ ways, and the boys in $4!$.
Remark: The problem with deciding that we will divide by something is that then we are doing manipulation and not visualization. I prefer to keep things concrete, to retain control over the counting process.