 Solution

Put $A=\mathcal{O}(X)$ and $\neg a=a\to 0$. The closure of the open $a$ is the closed sublocale with closed nucleus $c(\neg a)$; its interior is the open $\neg \neg a$. The closed sublocale complementary to $\neg a$ is also open precisely when $\neg a$ is complemented, in which case its open representative is $\neg \neg a$. Thus the closure is clopen precisely when the regular open element $\neg \neg a$ is complemented. Every regular open element has this form, so (i) and (ii) are equivalent.