Solution:

a) Let $O$ be the intersection of circles $ABD$ and $ACE$ (besides $A$). Since opposite angles in a cyclic quadrilateral sum to $180^\circ$, $\angle AOB = \angle AOC = 120^\circ$, so $\angle BOC = 120^\circ$ as well. Thus, quadrilateral $BOCF$ is also cyclic. Since $A$ and $O$ are the two intersection points of circles with centers $G$ and $H$, $AO \perp GH$. Similarly, $BO \perp GI$. Since $\angle AOB = 120^\circ$, $\angle GHI = 60^\circ$. Similarly, $\angle HGI = \angle HIG = 60^\circ$, so $\triangle GHI$ is equilateral.

b) We can use similar logic to show that $\triangle G'H'I'$ is equilateral as well (since $G'$, $H'$, and $I'$ are the centers of equilateral triangles drawn in the opposite direction on each side of $\triangle ABC$).

Next, we claim that
$$\triangle CHI' \cong \triangle CH'I \sim \triangle ABC.$$
We can see that $CH'$ is a $60^\circ$ rotation of $CH$ about point $C$, and $CI$ is a $60^\circ$ rotation of $CI'$ about $C$ in the same direction. Also, $\triangle ACH$ and $\triangle BCI$ are isosceles triangles with base angle $30^\circ$. Thus,
$$\begin{array}{r} CH = CH' = AC / \sqrt{3}, \\ CI' = CI = BC / \sqrt{3}, \\ \angle HCI' = \angle ACB = \angle H'CI. \end{array}$$
The desired similarity and congruence follows by SAS, and we conclude that
$$IH' = I'H = AB / \sqrt{3} = GG'.$$
By similar logic,
$$HH' = IG' = G'I.$$
Thus by SSS,
$$\triangle IHH' \cong \triangle HI'G \cong \triangle GG'I.$$
Thus, the vertices of $\triangle G'H'I'$ are positioned symmetrically with respect to the sides of $\triangle GHI$, and it follows that the two equilateral triangles have the same center.