Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

The idea is to first make the Earthlings sit, then make the Martians sit in the gaps between two Earthlings.

a)

There are $n!$ ways in which the Earthlings can sit. After they sit, there are $n+1$ ($n-1$ gaps between two Earthlings and $2$ positions at the sides) empty positions for the Martians to sit. There are $^{n+1}P_{m}=m!\binom{n+1}{m}$ ways for the $m$ Martians to sit in $n+1$ positions. Therefore, the answer is $\boxed{m!\cdot n!\cdot\tbinom{n+1}{m}}$.

b)

There are $(n-1)!$ ways for the Earthlings to sit around a round-table. But this time, there are $n$ positions for the Martians to sit. There are $^{n}P_{m}=m!\binom{n}{m}$ ways in which the Martians can sit. Hence, the total number of ways is $\boxed{m!\cdot(n-1)!\cdot\tbinom{n}{m}}$.