Graph theory, $n$ people sitting around table
Point 3. If the seats are unlabelled, the number of arrangements is $\frac{(n-1)!}{2}$.

Explanation: put a hat on one of the person and take it as the leader. Then you have $n-1$ choices for the person to his right and so on. But each arrangement is counted twice if you consider that circular arrangement $(a,b,c,d)$ is the same as the circular arrangement $(a,d,c,b)$.

Point 2. Answer is no. Take four people. They can only share one meal in your conditions. The people facing each-other will never eat side by side :-(

Point 1. You basically want to partition the edges of the complete graph on $n$ vertices into $k$ hamiltonian cycles. An obvious upper bound is then that $nk \leq {n \choose 2}$ so that $k \leq \left\lfloor \frac{n-1}{2} \right\rfloor$. This upper bound is met when $n$ is a prime number (take a first cycle then people at distance 2 then people at distance 3...)