a. Let $z_1, z_2, \cdots, z_n$ be complex numbers corresponding to the points. Let $S = \{z_1, z_2, \cdots z_n\}.$ Consider the sets $S_i = \{z_i + x| x \in S\}.$ These sets correspond to copies of $A$, and they work. b. Here we consider the sets $S_i = \{(p_i+c) \cdot (x+c) | x \in S \}$. We will pick suitable $c \in \mathbb{C}$ so that these sets work. First of all notice that any two sets share at least one point. We will now ensure that no other points are shared between sets. To do this, we need to make sure that $(p_x + c)(p_y + c) \neq (p_z + c)(p_w+ c)$ for pairs $(x, y), (z, w)$ so that $\{x, y\} \neq \{z, w\}.$ For any $x, y, z, w$ so that $p_x + p_y \neq p_z + p_w$, it's clear that the number of $c$ with $(p_x + c)(p_y + c) = (p_z + c)(p_w+ c)$ is precisely one. If $p_x + p_y = p_z + p_w$, it remains to note that $p_x p_y \neq p_z + p_w$, else Vieta implies $\{p_x, p_y\} = \{p_z, p_w\} \Rightarrow \{x, y\} = \{z, w\}$. Hence, there are only finitely many $c$ so that $(p_x + c)(p_y + c) = (p_z + c)(p_w+ c)$ for some $x, y, z, w$ so that $\{x, y\} \neq \{z, w\}$, and so we are done as long as we take any $c$ which is not these. $\square$