$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$Your argument for (i) doesn’t work, because it’s not true in general that $A=\cl\int A$ when $A$ is a closed set. For example, suppose that $A$ is the middle-thirds Cantor set in $\Bbb R$; then $A$ is closed, but $\int A=\varnothing$, so $\cl\int A=\varnothing\ne A$. For that matter, just let $A=\{0\}$: this is also a closed set in $\Bbb R$, but $\cl\int\{0\}=\cl\varnothing=\varnothing\ne\{0\}$.
Let $A$ be a closed set in a space $X$, and let $U=\int A$. Clearly $U\subseteq A$, so $\cl U\subseteq\cl A=A$, and therefore $\int\cl U\subseteq\int A=U$. It only remains to show that $U\subseteq\int\cl U$. HINT: Start from the fact inclusion $U\subseteq\cl U$ and take interiors on both sides.
For (ii) you can argue in very similar fashion. Start with an open set $U$, and let $A=\cl U$. Then $U\subseteq A$, so $U=\int U\subseteq\int A$, and ... ?
Alternatively, you can start with an open set $U$ and let $A=X\setminus U$. Then $A$ is closed, so by (i) we know that $\int A$ is regular open. If you know that the complement of a regular open set is regular closed, you can conclude that $X\setminus\int A$ is regular closed and try to show that it’s equal to $\cl U$.