Let $F(n)$ be the number of ways of seating $n$ people. We could start by carefully listing the ways, for a few small values of $n$. The listing should be almost explicit. We find that $F(1)=1$, $F(2)=2$, $F(3)=6$, and (perhaps) that $F(4)=24$. Despite the small amount of evidence, the conjecture $F(n)=n!$ is tempting.
We prove the conjecture, in principle by induction. Suppose that we know that for a specific $k$, $F(k)=k!$. Now George, the $(k+1)$-th person, comes along, late as usual. For every possible seating of the $k$ people, we can place George at the table of one of the $k$ people, and immediately to the right of that person ($k$ choices), or we can place George at a table by himself. Thus
$$F(k+1)=kF(k)+F(k)=(k+1)F(k)=(k+1)!.$$
Since $F(1)=1$, we conclude that $F(n)=n!$ for all $n$.
Another way: We can use fancier language. For example, note that every permutation of the set $\{1,2,\dots,n\}$ can be expressed uniquely as a product of disjoint cycles. We explicitly include any $1$-cycles. The order in which the product is taken does not matter, since the cycles are disjoint.
Every product of disjoint cycles corresponds to a unique circular seating, and vice-versa. (The people in a cycle determine a table, and their cyclic order determines the order of seating at that table.) This gives an explicit bijection between the permutations of $\{1,2,\dots,n\}$ and the seatings. Thus the number of seatings is equal to $n!$, the number of permutations.
The exclusion of arrangements that can be obtained from rotation comes to the same as the extra condition that $A$ is seated e.g. at the upper side. This because in any case there is exactly one rotation that brings him there. Then there are
$2$ possibilities for $A$. The first part then gives $2\times7!=10080$
possibilities, confirming your own answer. The second part gives $2\times6\times6!=8640$
possibilities (if I understand well that $A$ and $B$ are not sitting
next to eachother here). The factor $6$ corresponds with the possibilities
for $B$.
Best Answer
It's not necessary to count the number of arrangements around either table; the ratio is simply $3$, because each of the equilateral arrangements corresponds to three isosceles arrangements, letting the "bases" of the isosceles triangles be the three different sides of the equilateral triangle. (This assumes "isosceles" means "isosceles but not equilateral.")
Remark: I was initially just going to post this as a comment, but decided to make it a full fledged answer to stress a point: If you play close attention to exactly what a question asks for, you can sometimes get away with doing a lot less work than you might otherwise anticipate. In this case, actually counting the numbers of arrangements isn't terribly difficult, but it does take some effort. Since the question only asks for the ratio, it suffices to observe that the isosceles arrangements come in groups of three. A more advanced problem might pose a daunting challenge to count the arrangements but still have a simple way to arrive at the ratio.