Let $\times$ represents a girl and $o$ represents a boy.
Then, first place the girls as $ .\times . \times.\times.\times.\times.\times.$ where $.$ represents $o$ or no one. Now , Since John is at extreme, he has two choices while other $4$ boys have $6$ choices giving total of ${2\choose 1}{6\choose 4}$. Now permute all the boys independently(except John) and girls independently giving total of ${2\choose 1}{6\choose 4}4!6!$ sitting arragements.
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
First seat the $4$ girls around the table, which can be done in $(4-1)!$ ways (why?, Check "circular permutation"). Now in between these girls, there are $4$ gaps, where you would have to seat the $5$ boys, two of whom must sit together. So we club the two boys to form a single unit and see that the $4$ units can be seated in the $4$ gaps in $4!$ ways. However the two boys who form the unit, can be rearranged among themselves in $2$ ways. So total number of ways is $2!\times3!\times4!$.