Let $T$ be a fixed theory (recursively axiomatized, extending $I\Delta_0+\mathrm{EXP}$, sound). I read your first question as:

Q1: Is there a sentence $A$ such that $T$ proves that “$T$ does not prove ‘$T$ proves $A$ or $T$ proves $\neg A$’ and $T$ does not prove ‘$T$ does not prove $A$ and $T$ does not prove $\neg A$’”?

The answer is no, by Gödel’s theorem: if $T$ does not prove some formula, then in particular $T$ is consistent, hence a positive answer would imply that $T$ proves its own consistency. In light of this argument, we can make the question more sensible by putting enough consistency of $T$ in the assumptions:

Q1’: Is there a sentence $A$ such that $T$ proves that “If $T$ + ‘$T$ is consistent’ is consistent, then $T$ does not prove ‘$T$ proves $A$ or $T$ proves $\neg A$’ and $T$ does not prove ‘if $T$ is consistent, then $T$ does not prove $A$ and $T$ does not prove $\neg A$’”?

The answer is yes, and the convenient way to solve this and similar question is to use provability logic.

The basic provability logic works with the language of propositional modal logic: we have propositional variables $p_0$, $p_1$, $p_2$, ..., Boolean connectives such as $\land$, $\lor$ and $\neg$ (including the constants $\bot$ and $\top$ for falsity and truth), and the unary modal connective $\Box$. We also define $\Diamond A=\neg\Box\neg A$. An arithmetical interpretation of this language is an assignment $*$ of an arithmetical sentence $p^*$ to every propositional variable $p$, which is extended to all modal formulas by making it commute with Boolean connectives, and putting $(\Box A)^*=\Pr_T(\ulcorner A^*\urcorner)$, where $\Pr_T$ is the formalized provability predicate for $T$. That is, we read $\Box A$ as “$A$ is provable in $T$”, and $\Diamond A$ as “$A$ is consistent with $T$”. In particular, $\Diamond\top$ translates to “$T$ is consistent”.

$\DeclareMathOperator\prl{PRL}$
Then, the provability logic of $T$ with respect to a metatheory $S$, denoted $\prl_S(T)$, is the set of all modal formulas $A$ such that $S\vdash A^*$ for every arithmetical interpretation $*$. The two most important special cases are the ordinary provability logic of $T$, which is $\prl(T):=\prl_T(T)$, and the true provability logic of $T$, which is $\prl^+(T):=\prl_{\mathrm{Th}(\mathbb N)}(T)$ (here, $\mathrm{Th}(\mathbb N)$ is the true arithmetic, i.e., the set of all arithmetical sentences true in the standard model).

The modal logics $\prl_S(T)$ have been completely characterized for all sufficiently strong theories $T,S$. In particular, under our assumptions, $\prl(T)$ is the so-called Gödel–Löb logic GL, and $\prl^+(T)$ is Solovay’s logic S. Both logics are decidable, and they have transparent Kripke semantics.

Now, a moment’s reflection shows that Q1’ is equivalent to:

Q1’’: the formula $B:=\neg\Box[\Diamond\Diamond\top\to\neg\Box(\Box p\lor\Box\neg p)\land\neg\Box(\Diamond\top\to\neg\Box p\land\neg\Box\neg p)]$ is not in $\prl^+(T)$.

One can easily produce a Kripke model of S where $B$ fails, hence the answer is indeed positive. One can solve all sorts of similar questions about unprovability like this, by expressing it in provability logic and then finding a suitable Kripke model. You can learn more about provability logic from references given in the Wikipedia article.