If Cov(A,B) = 0, what can be said about Cov(|A|, |B|)? If I have two random variables $A$ and $B$ taking values in $[-1,1]$ (where both $1$ and $-1$ have some non-zero weight), and I know that $Cov(A,B) = 0$, can anything at all be said about $Cov(|A|, |B|)$?
If not, would anyone possibly be able to give me an example where $Cov(A,B)=0$, but $Cov(|A|,|B|)$ can be tuned arbitrarily?

Yes, it is possible.
Let $A=\pm 1$ with probabilities $1/4$ or $0$ with probability $1/2$, $Z$ be symmetric Bernolli, and $B = AZ.$ Then $\operatorname{cov} (A,B) = 0$, but $\operatorname{cov}(|A|,|B|) = \frac{1}{4}$. (This is the maximal possible value under these conditions.) Further, let $B' = Z\mathbf{1}_{A=0}$. Then $\operatorname{cov} (A,B') = 0$, but $\operatorname{cov}(|A|,|B'|) = -\frac{1}{4}$. (This is the minimal possible value under these conditions.)
Everything in between may be achieved by mixing these two examples.
