Your question amounts to prove that the determinant of the matrix $A-\lambda E$ has the form:
$$ p(\lambda) = (-1)^n\lambda^n+E_1\lambda^{n-1}+E_2\lambda^{n-2}+ \dots +E_{n-1}\lambda + E_n$$
Please notice the $(-1)^n$ sign in front of the first $\lambda$ that is originally missing in your question but which is crucial for the theorem to be proved.
As correctly suggested by Horn and Johnson, the above can be proved by using mathematical induction (on the size of square matrix A) applying the Laplace expansion in the inductive step as follows:
Base case $size(A)=2$ (we omit $n=1$ being trivial)
Consider the determinant of an $2x2$ matrix $\det(A-\lambda E)$
$$
\begin{vmatrix}
a-\lambda & b \\
d & e-\lambda \\
\end{vmatrix}
$$
If we consider $\det(A-\lambda E)$ as follows
$$
\begin{vmatrix}
a-\lambda & b+0\\
d+0 & e-\lambda \\
\end{vmatrix}
$$
we can see each column of $\det(A-\lambda E)$ as a linear combination of two columns of numbers and, therefore, we can apply the linear property of determinants by "separating out the powers of $\lambda$" (as illustrated in https://math.stackexchange.com/a/144130/557814) by obtaining:
$$
\begin{vmatrix} a-\lambda & b \\
c & d-\lambda \end{vmatrix} =
\begin{vmatrix} a & b \\
c & d-\lambda \end{vmatrix} +
\begin{vmatrix} -\lambda & b \\
0 & d-\lambda \end{vmatrix} =
\begin{vmatrix} a & b \\
c & d \end{vmatrix} + %%
\begin{vmatrix} a & 0 \\
c & -\lambda \end{vmatrix} + %%
\begin{vmatrix} -\lambda & b \\
0 & d \end{vmatrix} +
\begin{vmatrix} -\lambda & 0 \\
0 & -\lambda \end{vmatrix} = \\
(-1)^2\lambda^2 +E_1\lambda{2-1}+E_2
$$
Inductive step (size(A) $n$ assuming the for matrixes with size $n-1$ the hypothesis is true)
Consider the following determinant:
$$
\begin{vmatrix}
a_{1,1}-\lambda & a_{1,2} & \dots & a_{1,n-1} & a_{1,n} \\
a_{2,1} & a_{2,2}-\lambda & \dots & a_{2,n-1} & a_{2,n} \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1}-\lambda & a_{n-1,n} \\
a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & a_{n,n} - \lambda \\
\end{vmatrix}
$$
Now, by applying the linear property of determinants (as done the base case) we have:
$$
\begin{vmatrix}
a_{1,1}-\lambda & a_{1,2} & \dots & a_{1,n-1} & a_{1,n} \\
a_{2,1} & a_{2,2}-\lambda & \dots & a_{2,n-1} & a_{2,n} \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1}-\lambda & a_{n-1,n} \\
a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & a_{n,n} - \lambda \\
\end{vmatrix} = \\
\begin{vmatrix}
a_{1,1} & a_{1,2} & \dots & a_{1,n-1} & a_{1,n} \\
a_{2,1} & a_{2,2} & \dots & a_{2,n-1} & a_{2,n} \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1} & a_{n-1,n} \\
a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & a_{n,n}\\
\end{vmatrix} +
\begin{vmatrix}
a_{1,1}-\lambda & a_{1,2} & \dots & a_{1,n-1} & 0 \\
a_{2,1} & a_{2,2}-\lambda & \dots & a_{2,n-1} & 0 \\
\vdots & \vdots & \dots & \vdots & \vdots \\
a_{n-1,1} & a_{n-1,2} & \dots & a_{n-1,n-1}-\lambda & 0 \\
a_{n,1} & a_{n,2} & \dots & a_{n,n-1} & - \lambda \\
\end{vmatrix}
$$
Now, by applying the Laplace expansion on the last column of the second determinant in the above formula we obtain:
$$
\det(A)
-\lambda \det(M_{n,n}-\lambda E)
$$
where $M_{n,n}$ is the matrix that results by deleting the $n$th row and column from $A$. But the above, by the inductive hypothesis equals to:
$$
\det(A) -\lambda[(-1)^{n-1}\lambda^{n-1}+E_1\lambda^{(n-1)-1}+E_2\lambda^{(n-1)-2}+\dots+E_{n-1})] = \\ (-1)^{n}\lambda^n +E_1\lambda^{n-1}+E_2\lambda^{n-2}+\dots+E_{n-1}\lambda+E_n = p(\lambda)
$$
QED
Best Answer
I can help you with part 2 for now; part 4 involves too much terminology I have forgotten or never learned.
Just as Lombardi and Quitté do, I suppose that $a = 0$, since the general case can be transformed into this one by replacing $A$ by $A - aI_n$. Thus, the characteristic polynomial $f$ of $A$ admits $0$ as a simple zero, and we have $g = f/X$ and $h=X$ and $K = \operatorname{Ker} A$ and $I = \operatorname{Im} A$.
My impression is that the Schur complement will do the trick for part 4, but I'd have to look up the exact definition of "simple".