Recall that a necessary and sufficient condition for $k \ge 3$ positive real numbers $s_1, \dots, s_k$ to be the side lengths of a non-degenerate planar $k$-gon is that a maximal $s_i$ be less than the sum of the other $s_j$.

a) The answer is in the affirmative. Given pairwise distinct positive real numbers $\varepsilon_1, \varepsilon_2, \varepsilon_3$ less than $1/2$, we show that the sets $S = \{1, 2, 4, \dots, 2^{n-5}, 2^{n-4} + \varepsilon_1, 2^{n-4} + \varepsilon_2, 2^{n-4} + \varepsilon_3, 2^{n-3} - 1/2\}$ and $S \setminus \{2^{n-3} - 1/2\}$ are both multipolygonal.

Split any of the two sets into two subsets each of which has at least three elements, let $A$ be the part containing at least two of the $2^{n-4} + \varepsilon_i$, and let $B$ be the other part.
The set $A$ is polygonal since its maximal element is either one of the $2^{n-4} + \varepsilon_i$ or $2^{n-3} - 1/2$, each of which is smaller than the sum of other elements in $A$.
To prove that $B$ is not polygonal, notice that its maximal element is either $2^{n-3} - 1/2$, or one of the $2^{n-4} + \varepsilon_i$, or some $2^k$, $k \le n-5$. In the first case, the sum of all other elements in $B$ is less than $1+2+\dots+2^{n-5}+2^{n-4}+\varepsilon_i = 2^{n-3}-1+\varepsilon_i < 2^{n-3}-1/2$; in the second case, this sum does not exceed $1+2+\dots+2^{n-5} = 2^{n-4}-1 < 2^{n-4}+\varepsilon_i$; and in the third case, this sum is at most $1+2+\dots+2^{k-1} = 2^k - 1 < 2^k$. Consequently, $B$ is not polygonal.