(2) As you observe, the characteristic polynomial of $A$ is $x^4$, so it has sole eigenvalue zero (of multiplicity $4$). Now, if we denote by $J_k$ the $k \times k$ Jordan block with eigenvalue $0$, the possible Jordan forms of such a matrix are
$$J_4, \quad J_3 \oplus J_1, \quad J_2 \oplus J_2, \quad J_2 \oplus J_1 \oplus J_1, \quad 0.$$
Now, $$\ker A = \langle e_4 \rangle,$$
and in particular $\dim \ker A = 1$, but the only Jordan-form matrix above with $1$-dimensional kernel is $J_4$, and the dimension of the kernel of two similar matrices agrees, so $J_4$ must be the Jordan form of $A$. In fact, as abel observes, the number of Jordan blocks in the Jordan form a matrix with all zero eigenvalues (a nilpotent matrix) is exactly the dimension of its kernel.
(1) Jordan blocks $J_k$ of eigenvalue zero have the property that $J_k^r \neq 0$ for $r < k$ and $J_k^r = 0$ for $r \geq k$. So, $J_4^3 \neq 0$ and hence $A^3 \neq 0$, that is, the minimal polynomial of $A$ must be $x^3$. More generally, the minimal polynomial of a matrix with all zero eigenvalues is $x^r$, where $r$ is the size of the largest Jordan block in its Jordan form.