$AB*\text{adjoint}(BA)=I$ $AB*\text{adj}(BA)=I$ Prove: $1$. $|AB|=1$ $2$. $AB=BA$ As for $2$. what I have menage is $AB*AB^{-1}=AB^{-1}*AB=AB*\text{adj}$(BA)=I$ \rightarrow BA=AB$ How do I solve $1$. and is $2$. is valid?

From your assumptions it follows that $A$, $B$, and hence also $\text{adj}(AB)$ and $\text{adj}(BA)$ are invertible. Just apply determinant to the equation $AB\cdot\text{adj}(BA)=I$, and use the fact $M\text{adj}M=\det(M)I$.

Moreover, the assumption implies $$ (\text{adj}(BA))^{-1} = AB. $$ On the other hand using $M\text{adj}M=\det(M)I$ we find $$ (\text{adj}(BA))^{-1} = (\det(BA))^{-1} BA $$ This implies $$ \det(BA)AB = BA. $$ Applying determinant to this equation yields $\det(BA)=1$, which gives and $AB=BA$.