Recall that $I\Delta_0$ is the theory in the language of arithmetic that consists of the axioms of $\mathsf{PA}$ with induction restricted to $\Delta_0$ formulas (i.e., formulas where all quantifiers are bounded).

It is not too difficult to build a model $M$ of $I\Delta_0$ with a proper cut $N \subset M$ such that $M \equiv N$. Moreover, we can ensure that $2^n$ exists for every $n \in N$ and that $2^x$ is not a total function on $N$ (and therefore not a total function on $M$ either). It is also a standard fact that the set $\{ x \in M : (\exists y \in N) x < 2^y\}$ is a cut that models $I\Delta_0$ (this follows from the fact that it is downwards closed and closed under addition and multiplication). My question is about this last cut actually being all of $M$.

Let $C$ be a unary predicate symbol, and let $T$ be the theory in the language of arithmetic augmented with $C$ that contains $I\Delta_0$ and says that

$C$ is downwards closed,

$C$ is not all of the structure,

$2^x$ is not a total function,

$2^x$ is total on $C$,

the induced structure on $C$ has the same theory as the full structure (in particular, it is a model of $I\Delta_0$), and

for every $x$, there exists a $y \in C$ such that $x < 2^y$.

Obviously, the fifth bullet point must be stated as an axiom scheme.

Question 1.Is $T$ consistent?

Question 2.Does $I\Delta_0$ interpret $T$?

Question 3.Does $T$ interpret $I\Delta_0 + \mathrm{Exp}$ (where $\mathrm{Exp}$ is the statement that $2^x$ is a total function)?

Note that since $I\Delta_0$ does not interpret $I\Delta_0 + \mathrm{Exp}$, these last two questions cannot both have a positive answer.