$\newcommand{\cl}{\operatorname{cl}}$I’ll do part of it and leave part as hints.
It should be very easy to prove that $(1)$ and $(3)$ are equivalent. It’s also pretty easy to prove that $(1)$ implies $(5)$: if $A$ is dense in $X$, then $\cl A=X$, so ... ?
To prove that $(5)$ implies $(1)$, suppose that $A\subseteq X$; you want to show that $A$ is open in $\cl A$. Let $D=A\cup(X\setminus\cl A)$; show that $D$ is dense in $A$, and use the hypothesis $(5)$ to conclude that $D$ is open in $X$. From this you should have no trouble showing that $A$ is open in $\cl A$.
Once you’ve done all this, you’ll know that $(1),(3)$, and $(5)$ are equivalent. To show that these imply $(4)$, let $A$ be a subset of $X$, and let $D=\cl A\setminus A$; you want to show that $D$ is discrete. Show first that $X\setminus D$ is dense in $X$. Conclude that if $S\subseteq D$, then $X\setminus S$ is dense in $X$. Since you’re assuming the equivalent statements $(1),(3)$, and $(5)$, can then conclude that $X\setminus S$ is open for every $S\subseteq D$ and hence that every $S\subseteq D$ is closed in $X$. Then recall that if every subset of a set is closed, the set must be discrete.
It’s also not too hard to prove that $(5)$ is equivalent to $(2)$. Assume $(5)$. If $A$ is any subset of $X$, let $U=\operatorname{int}A$, the interior of $A$, and let $K=A\setminus U$. Certainly $U$ is open so we need only show that $K$ is closed. To do this, show that $X\setminus K$ is dense in $X$, hence open by $(5)$. Now assume $(2)$, and let $D$ be a dense subset of $X$. Let $A=X\setminus D$. By $(2)$ we can write $A=U\cup K$ for some open $U$ and closed $K$ in $X$. But $\operatorname{int}A=\varnothing$ (why?), so $U=\varnothing$, and $A=K$ is therefore closed, i.e., $D$ is open. Thus, $(1),(2),(3)$, and $(5)$ are equivalent, and they imply $(4)$.
I hve not yet suggested how to prove that $(4)$ implies $(1),(2),(3)$, and $(5)$, because this isn’t necessarily true for completely arbitrary topological spaces: the two-point space with the indiscrete topology satisfies $(4)$ but not $(5)$. Some additional hypothesis is necessary here.
Finally, for $(6)$ let $Y=\left\{\frac1n:n\in\Bbb Z^+\right\}$. If you find the right point $p\in\Bbb R$ and let $X=\{p\}\cup Y$ as a subspace of $\Bbb R$, you will have a space with the desired properties.
