Complement of NP-complete can be in NP
A problem $A$ is NP-complete if and only if $A'$ is in NP?


This is false. Take $A'=\emptyset$: then we have $A'\in\mathrm{NP}$ but $A=\Sigma^*$ is not $\mathrm{NP}$-complete.


  If there is an NP-complete language $L$ whose complement is in NP, then the complement of any language in NP is in NP.


This follows from the following three facts, which you should attempt to prove yourself.


The complement of an $\text{NP}$-complete problem is $\text{co-NP}$-complete;
if any $\text{co-NP}$-complete problem is in a class&nbsp;$\mathcal{C}$ that is closed under polynomial-time reductions, then every $\text{co-NP}\subseteq\mathcal{C}$;
$\text{NP}$ is closed under polynomial-time reductions.