Let the men be A, B, C, D. In problems about circular permutations, two permutations that differ by a rotation are usually considered to be the same. Equivalently, we can assume that there is a special chair, and that A is sitting on it.
Then the set of positions occupied by the men is determined, and the remaining men can be arranged in these in $3!$ ways. For each such way, the women can be arranged in $8!$ ways.
For the second problem, you will find the following approach useful. Call a placement bad if A and B are next to each other. Count the number of bad placements, and subtract this from the total number of possible placements.
Another way of solving the problem is to assume as before that A sits in the special chair. How many ways are there to seat B? And now how many ways are there to seat the rest?
Remark: It is sometimes a good idea to solve a problem in two different ways. That can provide a (partial) check of correctness.
In a row: There are $10!$ ways to arrange the people. We now count the bad arrangements, in which two A's are next to each other, or two B's, or two C's.
To do the counting, we use Inclusion/Exclusion. How many arrangements have the two A's together? Put them in a bag. We now have $8$ people and $1$ bag. These can be arranged in $9!$ ways. Then when we let the A's out of the bag, they can arrange themselves in $2$ ways, for a total of $2\cdot 9!$.
Do the same for the B's, the C's, and add up. We get $3\cdot 2\cdot 9!$.
However, we have double-counted the bad arrangements in which, for example, the A's and the B's are together. The same bag argument shows there are $2^2\cdot 8!$ such arrangements, for a total of $3\cdot 2^2\cdot 8!$.
Thus our next estimate for the number of bads is $3\cdot 2\cdot 9!-3\cdot 2^2\cdot 8!$.
However, we have subtracted one too many times the arrangements in which the A's are together, and the B's, and the C's. The now familiar technique shows there are $2^3\cdot 7!$ of these. So add back $2^3\cdot 7!$.
We end up with $7!(6\cdot 9\cdot 8-12\cdot 8+8)$, which is $8!\cdot 43$.
This is the number of bad arrangements. The number of good arrangements is $10!-8!\cdot 43$, which is $8!\cdot 47$.
Circular table: We use the convention that two orders are to be considered the same if they differ by a rotation. Assume that one of the "others" is a Canadian. He/she is probably too polite to complain, so one can put the Canadian in the worst chair.
The remaining chairs can now be thought of as a line. We have $9$ people, including $3$ "others" to put in a line, with the condition that the two A's, two B's, and two C's are separated. Same Inclusion/Exclusion technique.
Best Answer
The first problem is an inclusion-exclusion problem. Your $9!-2\cdot8!=357,120$ is the number of arrangements that have at least one of the pairs separated, not the number that have all three pairs separated.
The number of ways in which the two Americans sit together is $2\cdot 8!$: we treat them as a single individual, so we’re seating $9$ individuals, but that one individual has two ‘states’ that have to be distinguished, since the two can sit in either order. Similarly, there are $2\cdot 8!$ arrangements with the two British together and $2\cdot 8!$ with the two Chinese together. Thus, to a first approximation there are $9!-3\cdot2\cdot8!$ arrangements that have no two people of the same nationality seated together.
However, the figure $3\cdot2\cdot8!$ counts twice every arrangement that has both the Americans and the British together. There are $2^2\cdot7!$ such arrangements (why?), so we have to subtract this number from our original approximation. The same goes for the two other pairs, American and Chinese, and British and Chinese: in each of those cases we’ve also counted $2^2\cdot 7!$ arrangements in the figure of $3\cdot2\cdot8!$. Thus, our second approximation is $9!-3\cdot2\cdot8!+3\cdot2^2\cdot7!$.
Unfortunately, this still isn’t quite right: the $2^3\cdot6!$ arrangements that have the two Americans together, the two British together, and the two Chinese together were counted once each in the $9!$ term; subtracted $3$ times each in the $-3\cdot2\cdot8!$ term; and added back in $3$ times in the $3\cdot2^2\cdot7!$ term, so they have been counted a net total of one time each. But we don’t want to count them, so we have to subtract $1$ for each of them to reach the final answer:
$$\begin{align*} 9!-3\cdot2\cdot8!+3\cdot2^2\cdot7!-2^3\cdot6!&=(9-6)\cdot8!+(84-8)\cdot6!\\ &=3\cdot8!+76\cdot6!\\ &=244\cdot720\\ &=175,680\;. \end{align*}$$
To count the arrangements that have just the two Americans together, start with the $2\cdot8!$ arrangements that have them together. Among these there are $2^2\cdot7!$ that also have the two British together (why?), and $2^2\cdot7!$ that also have the two Chinese together. Finally, there are (as we saw before) $2^3\cdot6!$ arrangements that have all three pairs together. The inclusion-exclusion calculation is simpler this time:
$$2\cdot8!-2\cdot2^2\cdot7!+2^3\cdot6!=46,080\;.$$
I’ll leave the last one for you to try now that you’ve seen the first two.