Given
$$ \mathbf{M} =
\left(
\begin{array}{ccrc}
0 & 1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
0 & 7 & 0 & 6 \\
0 & 0 & 3 & 0 \\
\end{array}
\right),
$$
the determinant is nonzero:
$$
\det \mathbf{M} = 36
$$
No eigenvalues are $0$. This matrix has full rank. Both null spaces are trivial:
$$
\color{red}{\mathcal{N} \left( \mathbf{M} \right)} = \mathbf{0}, \qquad
\color{red}{\mathcal{N} \left( \mathbf{M}^{*} \right)} = \mathbf{0}
$$
Fundamental Theorem of Linear Algebra
Given a matrix
$$
\mathbf{M} \in \mathbb{R}^{m\times n}_{\rho}
$$
the four induced subspaces can be expressed as
$$
\begin{align}
%
\mathbb{R}^{n} =
\color{blue}{\mathcal{R} \left( \mathbf{M}^{*} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{M} \right)} \\
%
\mathbb{R}^{m} =
\color{blue}{\mathcal{R} \left( \mathbf{M} \right)} \oplus
\color{red} {\mathcal{N} \left( \mathbf{M}^{*} \right)}
%
\end{align}
$$
In this problem $m=n=\rho=4$, and there are no nullspaces. The two range spaces are
$$
\begin{align}
%
\text{Row }\mathbf{M} &=
\text{span} \left\{ \,
\color{blue}{
\left[
\begin{array}{c}
1 \\
0 \\
0 \\
0 \\
\end{array}
\right]},
\color{blue}{\left[
\begin{array}{c}
0 \\
1 \\
0 \\
0 \\
\end{array}
\right]},
\color{blue}{\left[
\begin{array}{c}
0 \\
0 \\
1 \\
0 \\
\end{array}
\right]},
\color{blue}{\left[
\begin{array}{c}
0 \\
0 \\
0 \\
1 \\
\end{array}
\right]}
\, \right\} = \mathbb{R}^{4}\\[5pt]
%
\text{Col }\mathbf{M} &=
\text{span} \left\{ \,
\color{blue}{
\left[
\begin{array}{c}
1 \\
0 \\
0 \\
0 \\
\end{array}
\right]},
\color{blue}{\left[
\begin{array}{c}
0 \\
1 \\
0 \\
0 \\
\end{array}
\right]},
\color{blue}{\left[
\begin{array}{c}
0 \\
0 \\
1 \\
0 \\
\end{array}
\right]},
\color{blue}{\left[
\begin{array}{c}
0 \\
0 \\
0 \\
1 \\
\end{array}
\right]}
\, \right\}
%
= \mathbb{R}^{4}
%
\end{align}
$$
(a) Null space basis
Both null spaces are trivial and are empty bases, as noted by @Omnomnomnom.
(b) Is $x$ in the null space?
Immediate answer: since the null space is trivial there are no null vectors such that $\color{red}{x}\in\color{red}{\mathcal{N} \left( \mathbf{M} \right)}$ such that
$$
\mathbf{M} \, \color{red}{x} = \mathbf{0}
$$
In fact,
$$
\mathbf{M} x =
\mathbf{M}
\left(
\begin{array}{c}
2 \\ 2 \\ 3 \\3
\end{array}
\right) =
\left(
\begin{array}{c}
2 \\ 1 \\ 32 \\ 9
\end{array}
\right)
$$
(c) Eigenvalues
The characteristic polynomial is defined as
$$
p(\lambda) = \det \left( \mathbf{M} - \lambda \mathbf{I}_{4} \right)
= \det
\left(
\begin{array}{rrrr}
-\lambda & 1 & 0 & 0 \\
2 & -\lambda & -1 & 0 \\
0 & 7 & -\lambda & 6 \\
0 & 0 & 3 & -\lambda \\
\end{array}
\right)
$$
The roots of the characteristic polynomial are the eigenvalues.
Attack using minors and cofactors. Level I:
$$
\det \left( \mathbf{M} - \lambda \mathbf{I}_{4} \right) =
- \lambda \det
\underbrace{\left(
\begin{array}{rrr}
-\lambda & -1 & 0 \\
7 & -\lambda & 6 \\
0 & 3 & -\lambda \\
\end{array}
\right)}_{\mathbf{S}}
-
\det
\underbrace{
\left(
\begin{array}{crc}
2 & -1 & 0 \\
0 & -\lambda & 6 \\
0 & 3 & -\lambda \\
\end{array}
\right)
}_{\mathbf{T}}
\tag{1}
$$
Level II:
$$
%
\begin{align}
%%
\det \mathbf{S} &=
\det
\left(
\begin{array}{rrr}
-\lambda & -1 & 0 \\
7 & -\lambda & 6 \\
0 & 3 & -\lambda \\
\end{array}
\right)
%
= -\lambda
\det
\left(
\begin{array}{rr}
-\lambda & 6 \\
3 & -\lambda \\
\end{array}
\right)
+
\det
\left(
\begin{array}{rr}
7 & 6 \\
0 & -\lambda \\
\end{array}
\right)
=
-\lambda^{3}+11\lambda \\[4pt]
%%
\det \mathbf{T} &=
\det
\left(
\begin{array}{crc}
2 & -1 & 0 \\
0 & -\lambda & 6 \\
0 & 3 & -\lambda \\
\end{array}
\right)%
= 2
\det
\left(
\begin{array}{rr}
-\lambda & 6 \\
3 & -\lambda \\
\end{array}
\right)
=
2\lambda^{2} - 36 \lambda
%
\end{align}
%
$$
Use these results in (1)
$$
\det \left( \mathbf{M} - \lambda \mathbf{I}_{4} \right) =
- \lambda \left( \lambda^{3} - 11\lambda \right)
- \left( 2\lambda^{2} - 36 \lambda \right)
$$
Characteristic polynomial:
$$
\boxed{
p(\lambda) = \lambda^4 - 13 \lambda^2 + 36
= \left( \lambda^{2} - 9 \right) \left( \lambda^{2} - 4 \right) =
\left( \lambda - 3 \right)
\left( \lambda + 3 \right)
\left( \lambda - 2 \right)
\left( \lambda + 2 \right)}
$$
Spectrum:
$$
\boxed{
\lambda\left( \mathbf{M} \right) = \left\{ \pm 3, \pm 2 \right\}
}
$$
(d) Eigenvectors
Given the matrix $\mathbf{M}$ and the eigenvalues $\lambda_{k}$ find the eigenvectors $v_{k}$ which solve the eigenvalue equation
$$
\mathbf{M} v_{k} = \lambda_{k} v_{k},
\qquad \left( \mathbf{M} - \lambda_{k} \mathbf{I}_{4} \right) v_{k} = \mathbf{0}
\quad k = 1,2,3,4
$$
The four eigenvectors follow.
$$
\begin{align}
\left( \mathbf{M} - (-3) \mathbf{I}_{4} \right) v_{1} &= \mathbf{0} \\
\left(
\begin{array}{rrrr}
3 & 1 & 0 & 0 \\
2 & 3 & -1 & 0 \\
0 & 7 & 3 & 6 \\
0 & 0 & 3 & 3 \\
\end{array}
\right)
v_{1} &=
\left(
\begin{array}{c}
0 \\ 0 \\ 0 \\ 0
\end{array}
\right)
\qquad \Rightarrow \qquad
v_{1} =
\left(
\begin{array}{r}
1 \\ -3 \\ -7 \\ 7
\end{array}
\right)
\end{align}
$$
$$
\begin{align}
\left( \mathbf{M} - (3) \mathbf{I}_{4} \right) v_{2} &= \mathbf{0} \\
\left(
\begin{array}{rrrr}
3 & 1 & 0 & 0 \\
2 & 3 & -1 & 0 \\
0 & 7 & 3 & 6 \\
0 & 0 & 3 & 3 \\
\end{array}
\right)
v_{2} &=
\left(
\begin{array}{c}
0 \\ 0 \\ 0 \\ 0
\end{array}
\right)
\qquad \Rightarrow \qquad
v_{2} =
\left(
\begin{array}{r}
1 \\ -3 \\ -7 \\ 7
\end{array}
\right)
\end{align}
$$
$$
\begin{align}
\left( \mathbf{M} - (-2) \mathbf{I}_{4} \right) v_{3} &= \mathbf{0} \\
\left(
\begin{array}{rrrr}
2 & 1 & 0 & 0 \\
2 & 2 & -1 & 0 \\
0 & 7 & 2 & 6 \\
0 & 0 & 3 & 2 \\
\end{array}
\right)
v_{3} &=
\left(
\begin{array}{c}
0 \\ 0 \\ 0 \\ 0
\end{array}
\right)
\qquad \Rightarrow \qquad
v_{3} =
\left(
\begin{array}{r}
1 \\ -2 \\ -2 \\ 3
\end{array}
\right)
\end{align}
$$
$$
\begin{align}
\left( \mathbf{M} - (2) \mathbf{I}_{4} \right) v_{4} &= \mathbf{0} \\
\left(
\begin{array}{rrrr}
-2 & 1 & 0 & 0 \\
2 & -2 & -1 & 0 \\
0 & 7 & -2 & 6 \\
0 & 0 & 3 & -2 \\
\end{array}
\right)v_{4} &=
\left(
\begin{array}{c}
0 \\ 0 \\ 0 \\ 0
\end{array}
\right)
\qquad \Rightarrow \qquad
v_{4} =
\left(
\begin{array}{r}
-1 \\ -2 \\ 2 \\ 3
\end{array}
\right)
\end{align}
$$
Examples of reductions for finding null spaces are Finding the basis of a null space, Deriving left nullspace of matrix from $EA=R$, Give bases for col(A) and null(A), Given a matrix and its reduced row echelon form, resolve the image and the kernel., Find base and dimension of given subspace
The matrix of eigenvectors is
$$
\mathbf{P} =
%
\left(
\begin{array}{cccc}
v_{1} &
v_{2} &
v_{3} &
v_{4}
\end{array}
\right)
%
=
%
\left(
\begin{array}{rrrr}
1 & -1 & 1 & -1 \\
-3 & -3 & -2 & -2 \\
-7 & 7 & -2 & 2 \\
7 & 7 & 3 & 3 \\
\end{array}
\right)
$$
To invert the matrix $\mathbf{P}$ use reduction and elementary row operations on the augmented matrix.
$$
\left(
\begin{array}{c|c}
\mathbf{P} & \mathbf{I}_{4}
\end{array}
\right)
\quad \rightarrow \quad
\left(
\begin{array}{c|c}
\mathbf{E_{P}} & \mathbf{P}^{-1}
\end{array}
\right)
$$
The first operation which clears column 1 is
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 \\
7 & 0 & 1 & 0 \\
-7 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr|cccc}
1 & -1 & 1 & -1 & 1 & 0 & 0 & 0 \\
-3 & -3 & -2 & -2 & 0 & 1 & 0 & 0 \\
-7 & 7 & -2 & 2 & 0 & 0 & 1 & 0 \\
7 & 7 & 3 & 3 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
%%
=
%%
\left(
\begin{array}{rrrr|rccc}
1 & -1 & 1 & -1 & 1 & 0 & 0 & 0 \\
0 & -6 & 1 & -5 & 3 & 1 & 0 & 0 \\
0 & 0 & 5 & -5 & 7 & 0 & 1 & 0 \\
0 & 14 & -4 & 10 & -7 & 0 & 0 & 1 \\
\end{array}
\right)
$$
The complete sequence of elementary matrices is clear columns 4, 3, 2, 1:
$$
%
\left(
\begin{array}{cccr}
1 & 0 & 0 & \frac{1}{5} \\
0 & 1 & 0 & \frac{1}{5} \\
0 & 0 & 1 & -\frac{3}{10} \\
0 & 0 & 0 & \frac{3}{10} \\
\end{array}
\right)
%
\left(
\begin{array}{ccrc}
1 & 0 & -\frac{1}{6} & 0 \\
0 & 1 & \frac{1}{30} & 0 \\
0 & 0 & \frac{1}{5} & 0 \\
0 & 0 & \frac{1}{3} & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{crcc}
1 & -\frac{1}{6} & 0 & 0 \\
0 & -\frac{1}{6} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & \frac{14}{6} & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{rccc}
1 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 \\
7 & 0 & 1 & 0 \\
-7 & 0 & 0 & 1 \\
\end{array}
\right)
$$
The above product is the inverse matrix:
$$
\mathbf{P}^{-1} =
\frac{1}{10}
\left(
\begin{array}{rrrr}
-2 & 3 & -1 & 2 \\
2 & 3 & 1 & 2 \\
7 & -7 & 1 & -3 \\
-7 & -7 & -1 & -3 \\
\end{array}
\right)
$$
$$
%
\boxed{
\begin{align}
%
%
\mathbf{P}^{-1} \, \mathbf{M} \, \mathbf{P} &= \mathbf{D} \\
%
\frac{1}{10}
\left(
\begin{array}{rrrr}
-2 & 3 & -1 & 2 \\
2 & 3 & 1 & 2 \\
7 & -7 & 1 & -3 \\
-7 & -7 & -1 & -3 \\
\end{array}
\right)
%
\left(
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
0 & 7 & 0 & 6 \\
0 & 0 & 3 & 0 \\
\end{array}
\right)
%
\left(
\begin{array}{rrrr}
1 & -1 & 1 & -1 \\
-3 & -3 & -2 & -2 \\
-7 & 7 & -2 & 2 \\
7 & 7 & 3 & 3 \\
\end{array}
\right)
%
&=
%
\left(
\begin{array}{rrrr}
-3 & \color{gray}{0} & \color{gray}{0} & \color{gray}{0} \\
\color{gray}{0} & 3 & \color{gray}{0} & \color{gray}{0} \\
\color{gray}{0} & \color{gray}{0} & -2 & \color{gray}{0}\\
\color{gray}{0} & \color{gray}{0} & \color{gray}{0} & 2 \\
\end{array}
\right)
%
\end{align}
}
%
$$
