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.
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.
Best Answer
Let $B_0$ and $G_0$ be the boy and girl who must not sit next to each other. Number the seats $1$ through $8$ from left to right. Suppose that $B_0$ sits in seat $1$. Then there are $3$ girls $-$ all of them except $G_0$ $-$ who can sit in seat $2$. After that the other $3$ boys can fill seats $3,5$, and $7$ in any order, and the other $3$ girls can fill seats $4,6$, and $8$ in any order, for a total of $3\cdot 3!\cdot 3! = 108$ ways to fill the seats with $B_0$ in seat $1$. An exactly similar calculation shows that there are also $108$ ways to fill the seats with $B_0$ in seat $8$.
If $B_0$ chooses to sit in one of the six seats in the middle, matters are a little different, because there are now two seats in which $G_0$ cannot sit. Note, though, that as soon as $B_0$ takes his seat, we know which $4$ seats must be taken by the girls, and we know which $3$ seats must be taken by the other $3$ boys. Those $3$ boys can sit as they please among those $3$ seats, so that’s $3!=6$ choices. $G_0$ has a choice of the $2$ ‘girl’ seats not adjacent to $B_0$: that’s another $2$ choices. Finally, the remaining $3$ girls can sit as they please among the remaining $3$ ‘girl’ seats, for another $3!=6$ choices. The grand total is then $6\cdot 2\cdot 6=72$ ways in which they can be seated once $B_0$ chooses a particular middle seat. Since there are $6$ middle seats, that comes to a total of $6\cdot 72 = 432$ ways.
Finally, combine the two cases: we have altogether $2\cdot 108 + 6\cdot 72 = 216+432=648$ ways to seat the $8$ of them.