Answer
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3 \\x_4 \end{aligned}\right]= \left[\begin{aligned}0\\0\\0\\0 \end{aligned}\right].$$
(b) The nullity is $0$.
(c) The rank of $A$ is $4$.
Since $A$ has $4$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=4+0=4.$$
Work Step by Step
Given the matrix
$$
A=\left[ \begin {array}{cccc} 1&2&1&2\\ 1&4&0&3
\\ -2&3&0&2\\ 1&2&6&1\end {array}
\right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&0
\\ 0&0&1&0\\ 0&0&0&1\end {array}
\right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} &=0\\
x_{2} &=0\\
x_{3} &=0\\
x_{4} &=0\\
\end{aligned}.
$$
The solution of the above system is $x_1= 0$, $x_2=0$,$x_3=0$, $x_4=0$.
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3 \\x_4 \end{aligned}\right]= \left[\begin{aligned}0\\0\\0\\0 \end{aligned}\right].$$
(b) The nullity is $0$.
(c) The rank of $A$ is $4$.
Since $A$ has $4$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=4+0=4.$$
