Option A
We will show that
\[
\mathrm{int}(X\setminus A) = X\setminus \overline {A}.
\]
In particular, the inclusion also holds. We recall that $\mathrm{int}(A) $ is the largest open set contained in $A$ and $\overline {A} $ is the smallest closed set containing $A$. Observe that
\begin{align*}
A \subseteq \overline {A} & \implies X \setminus \overline {A} \subseteq X\setminus A.
\end{align*}
As $\overline {A} $ is a closed set $X \setminus \overline {A} $ must be open and hence by definition of the interior of a set we have
\[
X \setminus \overline {A} \subseteq \mathrm{int}(X\setminus \overline {A} ).
\]
This proves one inclusion.
For the other inclusion, we observe that
\begin{align*}
A = X \setminus (X\setminus A) \subseteq X \setminus \mathrm{int}(X\setminus A).
\end{align*}
Since $\mathrm{int}(X\setminus A) $ is open, so $X\setminus \mathrm{int}(X\setminus A) $ must be closed and therefore, it must contain the closure of $A$. That is,
\begin{align*}
\overline {A} \subseteq X\setminus \mathrm{int}(X\setminus A) & \implies X \setminus \left( X\setminus \mathrm{int}(X\setminus A) \right) \subseteq X\setminus \overline {A} \\
& \implies \mathrm{int}(X\setminus A) \subseteq X\setminus \overline {A} .
\end{align*}
Therefore, we proved that
\[
\textcolor{teal}{\boxed{
\mathrm{int}(X\setminus A) = X\setminus \overline {A}
}}
\]
Option B
This is true just by the definition of the closure of a set. Closure of any set is the smallest set containing that set. If $A$ is closed, then $A = \overline {A} $.
Option C
This need not be true. We can choose a set $A$ whose interior is empty but boundary is nonempty. For example,
\[
X = \mathbb{R} \text{ and } A = \mathbb{Q}.
\]
Here we are considering the standard Euclidean topology. Then we have
\[
\mathrm{int}(A) = \emptyset \text{ and } \partial \mathbb{Q} = \overline {\mathbb{Q}} \setminus \mathrm{int}(\mathbb{Q}) = \mathbb{R} .
\]
But, the other inclusion is always true. To see this, note that
\begin{align*}
\mathrm{int}(A) \subseteq A \subseteq \overline {A} & \implies \overline{\mathrm{int}(A) } \subseteq \overline{\overline {A} } = \overline {A} \\
& \implies \overline{\mathrm{int}(A) } \setminus \mathrm{int}(A) \subseteq \overline {A} \setminus \mathrm{int}(A) \\
& \implies \partial (\mathrm{int}(A)) \subseteq \partial A.
\end{align*}
Option D
We have $\partial (\overline {A} ) = \overline{\overline {A} } \setminus \mathrm{int}(\overline {A} ) = \overline {A} - \mathrm{int}(\overline {A} )$. Since
\begin{align*}
\mathrm{int}(A) \subseteq A & \implies \overline {A} \setminus A \subseteq \overline {A} \setminus \mathrm{int}(A) \implies \partial (\overline {A} ) \subseteq \partial A.
\end{align*}