Step 1: Fix P at seat 0 (removes rotational repeats) and place Q at seat 2, exactly two seats to the right of P, using up 2 of the 8 seats.
Step 2: Count all ways to seat the remaining six people (R, S, T, U, V, W) in the 6 leftover seats while satisfying only the T–W gap-of-3 rule and the U–V opposite rule; direct checking gives 32 such layouts.
Step 3: Among those 32, remove the ones where S ends up next to R (violating the last rule). Exactly half of the 32, i.e. 16, have R and S adjacent, so they are discarded.
Step 4: What remains automatically satisfies every condition at once.
\[ 32 - 16 = \boxed{16} \]