Everywhere Super Dense Subset of $\mathbb{R}$

1) and 2): Yes. Take the union of countably many suitable "fat Cantor sets".

3) No. By the Lebesgue Density Theorem, if $A$ is a measurable subset of $\mathbb R$, $m(A \cap (t-\epsilon, t+\epsilon))/(2\epsilon) \to 1$ for almost every $t \in A$ and $0$ for almost every $t \notin A$.

EDIT with more details on 1) and 2):

Enumerate the open intervals $(a,b)$ with $a$ and $b$ rational and $a < b$ as $I_n = (a_n, b_n)$. I'll construct inductively disjoint sets $A$, $B$ as the union of sequences $A_n$, $B_n$ of closed nowhere-dense sets of positive measure (fat Cantor sets), where $A_n \subset I_n$ and $B_n \subset I_n$.

Given $A_1, \ldots, A_{n-1}, B_1, \ldots, B_{n-1}$, their union is a closed nowhere-dense set, so $I_n$ contains some interval $(c,d)$ disjoint from that set. Let $A_n$ be a fat Cantor set in $(c, (c+d)/2)$ and $B_n$ a fat Cantor set in $((c+d)/2, d)$.

Now let $A = \cup_{n=1}^\infty A_n$ and $B$ its complement (which contains $\cup_{n=1}^\infty B_n$). Any open interval $(a,b)$ contains some $I_n$ and its corresponding $A_n$ and $B_n$, and therefore $m((a,b) \cap A) \ge m(A_n) > 0$ and $m((a,b) \cap B) \ge m(B_n) > 0$.

EDIT: If $A$ is not measurable, you can't talk about the measure of its intersection with an interval. You could discuss outer measure, though: the (Lebesgue) outer measure of a set is the infimum of the measures of all Borel sets that contain it. You could, for example, consider a Bernstein set $A$. This has the property that every measurable set of positive measure intersects both $A$ and its complement. The result is that the outer measure
of $A \cap (a,b)$ and $A^c \cap (a,b)$ are both $b-a$.