What would be a not arithmetically definable language that is not Turing reducible to another given not arithmetically definable language? I have this question I'm struggling with. 

Let $A=\{&lt;i,n&gt;|\;n \in \phi ^{(i)}\}$. In other words, $A$ is the language defined by the set of all pairs $&lt;i,n&gt;$ such that $n$ is $\leq_m$ to the $i$th Turing jump.
A is definitely not an arithmetically definable language, and you can prove this by contradiction. 

I have to find a language, let's say $B$, that is more complex than $A$ $i.e.$ $B\nleq_T A$. I thought about $B=\{&lt;i,n&gt;|\;n \not\in \phi ^{(i)}\}$, but I'm not sure about my reasoning, I think an oracle turing machine with an oracle $A$ couldn't reduce to $B$..

And one last thing, do you think there exists a language that is more complex than any other (it doesn't reduce to anything)?

If you have any hints that could help me solve those two questions I would be really grateful!
Reiterating my comment, for each $A$ we can construct a language $A'$ (read: $A$ jump) which cannot be reduced to $A$ by taking all halting TMs with oracle access to $A$. The classical proof of the undecidability of the halting problem applies, showing that $A'$ cannot be (Turing-)reduced to $A$.