I understand what you are saying, but I think one problem is this: how do you know the inverse of a matrix is unique? It might not make sense to write $A^{-1}$ because this could represent multiple matrices. The fact that the inverse of a matrix is unique requires proof, at the very least. Here it is below:
Suppose $A$ is an $n \times n$ matrix and there exist $n \times n$ matrices $B$, $C$ such that $AB = BA = I$ and $AC = CA = I$. We want to show $B$ must equal $C$, and then that would mean the inverse of a matrix is unique (i.e., a matrix can't have two inverse matrices).
Well, $B = BI = B(AC) = (BA)C = IC = C$, and this shows that $B = C$. Note that I used the assumptions that $AC = I$ and $BA = I$, and also that matrix multiplication is associative (i.e., B(AC)=(AB)C).
So, now it makes sense to call the inverse of $A$ as $A^{-1}$, because we know if a matrix is invertible, by the above proof there is only one inverse.
Finally, as you said, we know if $A$ is invertible, then there is a unique matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. But this is precisely the statement we need to say $A^{-1}$ is invertible and its inverse is $A$. So, you are right. Once we know the inverse is always unique, the same statement that says $A$ is invertible also shows $A^{-1}$ is invertible and its inverse is $A$.
(a) Yes this is true, but you cannot think of a $4 \times 4$ matrix with a row of all zeroes as a $3 \times 4$ matrix. But the determinant of any matrix with an all zero row is zero, hence the matrix is not invertible.
(b) Correct
(c) follows straight from the definition. A matrix $A$ is said to be invertible if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. Note by this definition, $A^{-1}$ is also invertible with inverse $A$. To show that $A^2$ is invertible, just note that
$$A^2(A^{-1})^2 = AAA^{-1}A^{-1} = AIA^{-1} = AA^{-1} = I.$$
You can check that $(A^{-1})^2A^2 = I$ as well.
Best Answer
$\;AB\;$ invertible $\;\implies \exists\;C\;$ s.t.$\;C(AB)=I\;$ , but using associativity of matrix multiplication:
$$I=C(AB)=(CA)B\implies B\;\;\text{is invertible and}\;\;CA=B^{-1}$$