$7$ girls can be seated in $7!=5040$ ways. Each seating creates six inner slots where one or two boys can be placed, and two end slots where at most one boy can be placed.
When no two boys shall sit together we can place them in $8\cdot 7\cdot6=336$ ways in the eight slots ($8$ choices for the first boy, $7$ remaining for the second, and $6$ for the third).
When two boys shall sit together we can choose any of the $6$ inner slots for them. Then we can pick the lefthand one of the two in $3$ ways, the righthand one in $2$ ways. Finally we can choose any of the $7$ leftover slots for the third boy. In all we have $6\cdot3\cdot2\cdot 7=252$ possibilities for such an arrangement.
It follows that the total number of admissible seatings is given by
$$5040\cdot(336+252)=2\,963\,520\ .$$
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 the special boy takes a seat at his will. His choice then also determines which seats are for boys and which are for girls. He can choose an end seat in $2$ ways, leaving three admissible seats for the special girl, or an interior seat in $6$ ways, which leaves two admissible seats for the special girl. So there are $2\cdot 3+6\cdot 2=18$ ways to seat the special pair. The rest of the people can then be seated in $3!\cdot 3!=36$ ways on the remaining seats, making a total of $18\cdot36=648$ possibilities.