a. No. Let the centres be $A$, $B$, $C$, $D$, $E$. WLOG assume $A$ is the centre of $(BCD)$. One of the following must occur.

* If $B$ is the centre of $(ACD)$, then $AC = AD = AB = BC = BD$. So both $\triangle ABC$ and $\triangle ABD$ are equilateral triangles. Note that at least one of $A$, $B$ must lie on the circle with centre $C$. So $E$ lies on the circle with centre $C$ that passes through $A$ and $B$. Similarly, $E$ lies on the circle with centre $D$ that passes through $A$ and $B$. This is a contradiction since the two circles only meet at $A$ and $B$.

* If $B$ is the centre of $(ACE)$, then $AC = AD = AB = BC = BE$. So $\triangle ABC$ is an equilateral triangle. Thus, there must be $3$ points lying on the circle with centre $C$ that passes through $A$ and $B$. In either case $CA = CB = CD$ or $CA = CB = CE$, we get the same configuration as above and hence it is a contradiction.

b. Yes. Consider two equilateral triangles of side lengths $1$ having parallel sides and the same orientation such that each pair of corresponding vertices is at a distance $1$ apart. It is clear that each point has a distance $1$ to exactly three other points. Therefore, we can draw a circle with each point as centre and radius $1$ that passes through exactly $3$ other points.