In how many ways can eight people, denoted $A,B,C,D,E,F,G,H$ be seated about a square table that seats two people on each side?

My approach:

Since each side of the table seats two people, there are $2$ possible arrangements for two people sitting on a side (for example, $AB$ and $BA$). Given $8$ people, there are $8!$ ways to arrange them, and to avoid over counting rotations, we have $\frac{8!}{8}$ ways to arrange $8$ people around the table. So we have a total possible arrangements of
$$2\*\left(\frac{8!}{8}\right)=\frac{8!}{4}=10,080.$$

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If two of the eight people, say $A$ and $B$, do not get along well, how many different seatings are possible with $A$ and $B$ not sitting next to each other?

My approach:

There are two cases to consider by which $A$ is next to $B$, the first case $AB$ and the second case $BA$ (but they don't have to be on the same side of the square when this happens). These two cases give us $2\*6!$ possible arrangements.

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Is my approach to these problems correct? My approach to the second one feels incomplete, or even incorrect. As there must be a two step process, as I must find the number of arrangements by which the condition is met and also avoid over counting the rotations.