a) Assume that such a coloring exists and $a$, $b$ and $c$ are the colors. Let $O$ be the center of $k$ and consider an equilateral triangle $OBC$ with side $\sqrt{3}$. Its circumcircle has radius $1$ and touches $k$. If the color of $O$ is $a$, then the colors of $B$ and $C$ are $b$ and $c$. This shows that the points of the circle $k'(0, \sqrt{3})$ are colored in $b$ and $c$. Consider now an equilateral $\triangle PQR$ with circumcircle $k$. Denote by $X_1, X_2$ and $Y_1, Y_2$ the intersection points of $PQ$ and $PR$ with $k'$, respectively ($X_1$ and $Y_1$ the closer points to $P$). Since the circumcircle of the equilateral $\triangle PX_2Y_2$ touches $k$ and has radius at least $1$, it follows that the color of $P$ is $a$. Analogously $Q$ and $R$ have the same color, a contradiction.

b) Let $ABCDEF$ be a regular hexagon inscribed in $k$. Let the color of $O$ be $1$, let the color of the points inside the sector $OAB$, the radius $OA$ and the arc $AB$ without $B$ be $2$, let the color of the points inside the sector $OBC$, the radius $OB$ and the arc $BC$ without $C$ be $2$, etc. It is easy to see that this coloring has the desired properties.