Finding $|A|^2+|B|^2$ if $A=\text{adj} (B)-B^T$ and $B=\text{adj} (A)-A^T$ 
Consider two $3\times3$ matrices $A$ and $B$ satisfying $A=\text{adj} (B)-B^T$ and $B=\text{adj} (A)-A^T$ (where $C^T$ denotes transpose of matrix $C$). If $A$ is a non-singular matrix, then find $AB$ and $|A|^2+|B|^2$ (where $|C|$ denotes determinant of matrix $C$). Also, find $|\text{adj} (2B^{-1})|$.

Given, $A=\text{adj} (B)-B^T$. Taking transpose on both sides, I get $A^T=(\text{adj} B)^T-B$. Putting $B=\text{adj} (A)-A^T$ in it, I get, $A^T=(\text{adj}B)^T-\text{adj(A)}+A^T\implies\text{adj} A=(\text{adj} B)^T$. Taking determinant on both sides, I get $|A|^2=|B|^2$ (because $|\text{adj} C|=|C|^{n-1}$ where $n$ is the order of $C$ and $|C|=|C^T|)$.
Also, since $A$ is non-singular, if I divide $B=\text{adj} (A)-A^T$ by $|A|$, I get $\dfrac B{|A|} = A^{-1} - \dfrac{A^T}{|A|} \implies  \dfrac{B + A^T}{|A|} = A^{-1}$. Taking determinant on both sides, I get $\dfrac{|B+A^T|}{|A|^3} = \dfrac1{|A|} \implies |B + A^T| = |A|^2$ (because $|kC| = k^n|C|$, where $k$ is any constant (here $\frac1{|A|}$) and $n$ is order of matrix $C$. Also, $|C^{-1}|=\frac1{|C|}$).
Not able to proceed next.
EDIT: After @Upayan De's answer, $|B|=8$.
Now, $\text{adj}(kC)=k^{n-1}\text{adj}C$, where $k$ is a constant and $n$ is the order of matrix $C$.
So, $\text{adj}(2B^{-1})=4\text{adj}(B^{-1})$.
So, $|\text{adj}(2B^{-1})|=|4\text{adj}(B^{-1})|=64|\text{adj}(B^{-1})|$.
Now, $|\text{adj}(B^{-1})|=(|\text{adj}B|)^{-1}=(|B|^2)^{-1}=1/64$.
So, $|\text{adj}(2B^{-1})|=1$
 A: Taking transpose of the first equation, we have, $$A^T = (adj(B))^T - B$$ Substituting that in the second equation we have, $$\begin{aligned}B &= \text{adj}(A) - (\text{adj}(B))^T + B \\ (\text{adj}(B))^T &= \text{adj}(A) \\ \text{adj}(B^T) &= \text{adj}(A) \qquad \text{...(1)} \end{aligned}$$ Taking adjoint of both sides, and using this result, $$\begin{aligned} \text{adj(adj}(B^T)) &= \text{adj(adj}(B)) \\ |B^T|B^T &= |A|A\end{aligned}$$ Again, taking determinant of (1),$$|B^T|^2=|A|^2 \implies |B^T|=\pm|A|$$ Hence, we can conclude, $$B^T=\pm A$$
However, $A=-B^T$ on substituting in the first equation, gives $\text{adj}(B)=0 \implies |B|=0$
But $B$ is non-singular $\implies B^T=A$
Hence, from first equation, $$\begin{aligned}2A&=\text{adj}(A^T) \\ |2A| &= |\text{adj}(A^T)| \\ 8|A| &= |A^T|^2 \\ |A|&=|B|=8\end{aligned}$$ Therefore, $$|A|^2 + |B|^2 = 64 + 64 = 128$$
EDIT: My apologies for overlooking the other subparts of the question initially. Now since you have already worked out $|adj(2B^{-1})|$, I'll just post the answer AB
Post-multiplying first equation by $B$, we have, $$\begin{aligned}AB &= (\text{adj}(B))B - B^TB \\ AB &= |B|I - AB \\ 2AB &= 8I \\ AB &= 4I \end{aligned}$$ Hope that's all :-)
