This question already has an answer here:
For (1), if $rank(A) = n$, $adj(A)A = det(A)I$; taking determinants of both sides easily yields the result; if $rank(A) < n$, $det(A) = 0 = det(adj(A))$.
For (2), we use (1). If $rank A = n$, then
$$\begin{align}
adj(adj(A))adj(A) &= det(adj(A))I\\
&= det(adj(A)) (1/det(A))adj(A)A\\
&= det(adj(A))^{n-2} adj(A)A
adj(adj(A)) = det(adj(A))^{n-2}A,
\end{align}
$$
and, if $rank(A) < 0$, both sides are again zero.