Is there any solution?

Is there any solution?

(b) The color is uniquely determined by the initial coloring. Instead of red, white and blue color the vertices $0,1,2$ (respectively) notice that whenever we recolor, the sum of the labels over all vertices remains invariant modulo $3$. since $2005\equiv1(mod 3)$ the initial sum modulo $3$ is the last label.

Let $A,B,C$ be the colors. Observe that we can make $BAAB \mapsto BACC \mapsto BBBC$, $CAAB \mapsto CACC \mapsto CBBC \mapsto AAAA$, $BABA \mapsto CCCC, BAAA \mapsto CCAA \mapsto CBBA \mapsto CBCC \mapsto CAAC \mapsto BBBB$, $ABC \mapsto AAA$. The same is valid for cyclic positions. Hence, for any $4$ consecutive vertices, it's possible to turn the first three of them of the same color. Thus, do this operation, until we have many consecutive groups of $3$ consecutive vertices of the same color, and a single vertex $v$ with an unkwown color. Now, observe that we have the operation $AAAB \mapsto AACC \mapsto ABBC \mapsto CCBC \mapsto CAAC \mapsto BBBB$ (the same holds for cyclic positions). Hence, doing this operation many times clockwisely, starting at $v$, it's possible to make all the vertices with the same color as $v$, which completes item $(a)$. For item $(b)$, oberve that if at the beggining we have $x$ vertices of color $A$, $y$ vertices of color $B$ and $z$ vertices of color $C$, $x,y,z$ are not all equal $\pmod{3}$, since $2005 \equiv 1 \pmod{3}$. Assume WLOG $y \equiv z \pmod{3}$. If we make a movement switching $BC \mapsto AA$, the amount of vertices of colors $B$ and $C$ keeps the same $\pmod{3}$. If we make a movement switching $AB \mapsto CC$ , the amount of vertices of colors $B$ and $C$ keeps the same $\pmod{3}$, since $-1 \equiv 2 \pmod{3}$. The same is valid when we make $AC \mapsto BB$. Hence, the amount of vertices of colors $B$ and $C$ will always keep the same $\pmod{3}$. Hence, if at some point, the colors are all equal, it must be of color $A$, so we are done. $\blacksquare$