Let me get started with a small take at **question 3**. First, for $v\in\{1,2,3\}$ it is clear that no pooling design can have compression rate $r<1$ (so trivial is optimal). For example for $v=3$, we need to distinguish at least $5$ situations (no positives, at least $2$ positives, and $3$ possible single positives), so $2$ bits of information cannot be enough and we must have $e\ge 3$.

Thus $v=4$ is the first case where the trivial bound does not preclude a pooling design of interest (we need to distinguish $6$ situations, leading to the bound $e\ge3$). However:

**Proposition.**
There are no pooling design with $v=4$ and $r<1$.

*Proof.*
Assume $(V,E)$ is a pooling design with $V=\{1,2,3,4\}$ and $e=3$. If an element of $E$ is a singleton, then removing it from $E$ and its element from $V$ would give a pooling design with $v=3$ and $e=2$, which is impossible. If two elements $p,q$ of $E$ are contained one in the other, $p\subset q$, then replacing $q$ with $q\setminus p$ gives a pooling design (more information is carried by the results of $(p,q\setminus p)$ than by the results of $(p,q)$).
 
*We can thus assume that no element of $E$ is a singleton, and no element of $E$ contains another* (these are general arguments than can be used more widely).

In particular, all elements of $E$ have $2$ or $3$ elements.

*No vertex can belong to all edges*, since otherwise the positivity of this vertex would entail positivity of all edges, an event that cannot be distinguished from all vertices being positives. 

*No vertex $a$ can be contained in only one edge*, otherwise the positivity of another vertex $b$ of this edge could not be distinguished from the positivity of $a$ and $b$.

It follows that all vertices must have degree exactly $2$. The total degree is thus $8$, and we must have two elements of $E$ of cardinal $3$ and the last one of cardinal $2$. But then the two largest edges must have two elements in common, which thus have the same link, a contradiction.