It has a more difficult part : Prove that after any operations , the cards will not be in opposite order (counterclockwise) . This nice problem proposed by Dr.Naghshine , He has got the main idea from topology !! .

Unexpected!because It seems completely combinatorial (we have seen so many problems of this type),And Couldn't be solved without mahamath's hint! Anyway here is the unexpected solution: persons and cards are points.Suppose circumcenter is the origin .By segment $i$ I mean the segment between cards $i$ and $i+1$ and by triangle $i$ ,the triangle made by cards $i,i+1,i+2$.let $C$ be the closed directed path consisting of the 11 consecutive segments mentioned.let $s$ be the winding number of path around the origin and $s_{i}$ the value of the integral over segment $i$.we claim that $s$ is invariant .At first $s=-1$ and at last is $0$ in part (a) and $1$ in part (b). So both parts will be solved. Suppose in a step we have moved card $i$.By assumption the triangle $i-1$ won't contain origin before and after operation.let $X,Y,Z,T$ be respectively:card $i-1$,card $i$ before move,card $i$ after move,card $i+1$.By an illustration it will be easy to see that origin is outside closed path $XYTZ$ and so $s_{i-1}+s_{i}$ will be invariant in this step and so will be $s$.so we are done two notes: 1.If you imagine each step as a continuous move of a vertex along a line (and so segments and path will be transformed continuously) .if it can be proved that $C$ will never touch origin (*)then the rest can be done by topology/analysis .But if in a step that you move card $i$ , the cards $i-1$ and $i+1$ be on each other then (*) may not be true .So if we want to solve in this way we must think what to do when cards $i-1$ and $i+1$ are on each other 2.It can be shown that (b) implies (a) by combinatorial reasoning.