Rearrange the ACE icon with repeated symbols

There is one each of A,C,E,S,D, two F copies, two G copies and two asterisks: 11 symbols. Treat consecutive clockwise A,C,E as one block. This block has only one allowed internal order; do not multiply by 3! or 2!.

The ACE block,S,D,F,F,G,G are seven objects. Fix the unique ACE block as a circular reference and arrange the other six, correcting for two identical F copies and two identical G copies: $(7-1)!/(2!2!)=180$.

Those seven circular objects create seven gaps, including the gaps immediately before and after the ACE block but none inside it. Put the two identical asterisks in two different gaps, giving $\binom72=21$ choices. Total $[(7-1)!/(2!2!)]\binom72=180\times21=3780$.

Part (ii) is a separate condition; it does not retain the separation requirement of part (i). The ACE block and eight remaining symbols give nine outer objects. Fix ACE as a reference and correct for pairs of F,G and asterisks: $(9-1)!/(2!2!2!)=5040$ circular arrangements.

There are 11 choices for the labeled position occupied by A, which fixes the clockwise ACE block and anchors the circular order. Hence $[(9-1)!/(2!2!2!)]\times11=55{,}440$. Use 11 original symbol positions, not nine compressed objects; ACE occupies three consecutive positions and may wrap across the label boundary.