Step 1: Fix anchors using symmetry.
Place \(C\) at the top (without loss of generality). Since \(F\) is opposite \(C\), put \(F\) opposite \(C\).
Condition (iii) “\(H\) is the only one sitting between \(G\) and \(C\)” - the three consecutive seats are either \(G\!-\!H\!-\!C\) or \(C\!-\!H\!-\!G\).
Step 2: Try both orders; reject the impossible one.
If \(G\!-\!H\!-\!C\) (clockwise), then using (iv) \(E\) two to the right of \(B\) and “\(E\) not next to \(F\)” leads to a contradiction (no place for \(E\)).
Hence the only feasible order is \(C\!-\!H\!-\!G\) (clockwise).
Step 3: Place \(E\) and \(B\).
\(E\) is two to the right of \(B\) and not next to \(F\). The only way to satisfy both is \(E\) on the seat just before \(C\) (clockwise), which forces \(B\) two seats to its left (clockwise).
Step 4: Fill the remaining seats.
The last two seats go to \(A\) and \(D\), with \(D\) not next to \(F\) - \(D\) sits two to the right of \(F\).
Final circular order (clockwise): \(C,\;H,\;G,\;A,\;F,\;B,\;D,\;E\).
Here, \(H\) is directly opposite \(B\). \(\Rightarrow\) \(\boxed{\text{Exactly Opposite of B}}\).