In both cases, we can find such $A,B$. Let $k$ be sufficiently large integer. Then observe that for a regular $2k+1$ gon, with vertices $A_1A_2 \cdots A_{2k+1}$ inscribed in a circle of radius $R$, the ratio $\frac{A_1A_{k+1}}{R}$ can be made $> 2- \epsilon$ for any positive $\epsilon$. Since we know $\frac{\pi}{r} < 2$, we can choose an $R < r$ and $k$ such that $\frac{\pi}{r} = \frac{A_1A_{k+1}}{R}$. This means that $A_i, A_{k+i}$ must have opposite colors for all $i$, which is a contradiction since $2k+1$ is odd. $\blacksquare$

I will skip the details. Assume otherwise. If $O$ is the center of the disk, construct a circle $\omega$ with center $O$ radius $\frac{\pi}{2}<R<r$. Then for any close enough two points on $\omega$ we can find another point in disk which has the same distance $\pi$ to these points. Thus these two points have the same color. This implies all points $\omega$ has the same color. However we can find two points on $\omega$ with distance $\pi$. Contradiction.

Was the constant $\pi$ added to intimidate the contestants because I think we can scale down everything by $\pi$ so it holds no significance in the problem

For the first part, we can construct an equilateral triangle of side length $\pi$ on the circle of radius $\frac{\pi}{\sqrt{3}}$. For the second part, the answer is yes. We scale down the circle by a factor of $\frac{2}{\pi}$ for the sake of simplicity. Now, if we can construct a $2n+1$- pointed regular star of side length $2$ inside a circle of radius $r>1$, then we will be done (as at least one side of the star must have endpoints of same colour). Consider the $2n+1$-th roots of unity, the distance between $A_{n+1}(\cos (\frac{2n\pi}{2n+1}),\sin(\frac{2n\pi}{2n+1}))$ and $A_1(1,0)$ which tends to $2$ as $n$ grows larger, when $r=1$. So, for some points $X,Y$ on the segment $A_1A_{n+1}$, we will get $XY=2$, when $r>1$. Hence for some $n$ large enough, we can construct such $2n+1$-pointed regular star of side length $2$, and so we are done.