Solution:

(a) Given triangle $\triangle ABC$, Amy can draw the following points:
- $O$ is the circumcenter of $\triangle ABC$
- $A_1$ is the circumcenter of $\triangle BOC$
- $A_2$ is the circumcenter of $\triangle OBA_1$
- $A_3$ is the circumcenter of $\triangle BA_2A_1$

We claim that $AA_3 > 7$. We present two ways to prove this claim.

First Method:
By symmetry of the equilateral triangle $\triangle ABC$, we have $\angle AOB = \angle BOC = \angle COA = 120^\circ$. Since $OB = OC$ and $A_1B = A_1O = A_1C$, we deduce that $\triangle A_1OB \cong \triangle A_1OC$, and hence $\angle BOA_1 = \angle COA_1 = 60^\circ$. Therefore, since $\triangle A_1OB$ is isosceles, it must be equilateral. As we found for our original triangle, we find $\angle BA_2A_1 = 120^\circ$, and so $\angle A_2BA_1 = \angle A_2A_1B = 30^\circ$ (since $A_2B = A_2A_1$). Also we see that $\angle OBA_2 = 30^\circ = \angle OBC$, which shows that $A_2$ lies on the segment $BC$.

Applying the Law of Sines to $\triangle BOC$, we obtain
$$OC = \frac{BC \sin(\angle OBC)}{\sin(\angle BOC)} = \frac{6(1/2)}{\sqrt{3}/2} = 2\sqrt{3}.$$
By symmetry, we see that (i) $OA_1$ is the bisector of $\angle BOC$ and the perpendicular bisector of $BC$, and (ii) the three points $A$, $O$, and $A_1$ are collinear. Therefore $A_1A = A_1O + OA = 2OA = 4\sqrt{3}$.

The same argument that we used to show $\triangle A_1OB$ is equilateral with side $AC/\sqrt{3}$ shows that $\triangle A_3A_2A_1$ is equilateral with side $OB/\sqrt{3} = 2$. Thus $\angle A_3A_1O = \angle OA_1B + \angle A_3A_1A_2 - \angle A_2A_1B = 60^\circ + 60^\circ - 30^\circ = 90^\circ$. Hence we can apply the Pythagorean Theorem:
$$A_3A = \sqrt{(A_3A_1)^2 + (A_1A)^2} = \sqrt{2^2 + (4\sqrt{3})^2} = \sqrt{52} > \sqrt{49} = 7.$$

Second Method:
(An alternative to writing the justifications of the constructions in the First Method is to use analytic geometry. Once the following coordinates are found using the kind of reasoning in the First Method or by other means, the writeup can justify them succinctly by computing distances.)

Label $(0,0)$ as $B$, $(6,0)$ as $C$, and $(3, \sqrt{3})$ as $A$. Then we have $AB = BC = CA = 6$.

The circumcenter $O$ of $\triangle ABC$ is $(3, \sqrt{3})$; this can be verified by observing $OA = OB = OC = 2\sqrt{3}$. Next, the point $A_1 = (3, -\sqrt{3})$ satisfies $A_1O = A_1B = A_1C = 2\sqrt{3}$, so $A_1$ is the circumcenter of $\triangle BOC$.

The point $A_2 = (2, 0)$ satisfies $A_2O = A_2B = A_2A_1 = 2$, so this is the circumcenter of $\triangle OBA_1$.

And the point $A_3 = (1, -\sqrt{3})$ satisfies $A_3B = A_3A_2 = A_3A_1 = 2$, so this is the circumcenter of $\triangle BA_2A_1$.

Finally, we compute $A_3A = \sqrt{52} > \sqrt{49} = 7$, and part (a) is proved.

(b) In part (a), using either method we find that $OA_3 = 4 > 2\sqrt{3} = OA$. By rotating the construction of part (a) by $\pm 120^\circ$ about $O$, Amy can construct $B_3$ and $C_3$ such that $\triangle A_3B_3C_3$ is equilateral with circumcenter $O$ and circumradius $4$, which is strictly bigger than the circumradius $2\sqrt{3}$ of $\triangle ABC$. Amy can repeat this process starting from $\triangle A_3B_3C_3$. After $n$ iterations of the process, Amy will have drawn the vertices of an equilateral triangle whose circumradius is $2\sqrt{3}\left(\frac{4}{2\sqrt{3}}\right)^n$, which is bigger than $2019$ when $n$ is sufficiently large.