Answer
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}2t\\5t\\t\\t \end{aligned}\right] =t\left[\begin{aligned}2 \\5\\1\\1 \end{aligned}\right]+t\left[\begin{aligned}-1\\-2 \\0\\1 \end{aligned}\right].$$
(b) The nullity is $1$.
(c) The rank of $A$ is $3$.
Since $A$ has $4$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=3+1=4.$$
Work Step by Step
Given the matrix
$$
A=\left[ \begin {array}{cccc} 1&0&-2&0\\ 4&-2&4&-2
\\ -2&0&1&3\end {array} \right]
.
$$
The reduced row echelon form is
$$
\left[ \begin {array}{cccc} 1&0&0&-2\\ 0&1&0&-5
\\ 0&0&1&-1\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} -2x_4 &=0\\
x_{2} -5x_4 &=0\\
x_{3} - x_4 &=0
\end{aligned}.
$$
The solution of the above system is $x_1= 2t$, $x_2=5t$,$x_3=t$, $x_4=t$.
(a) The nullspace of $A $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2}\\x_3\\x_4 \end{aligned}\right]= \left[\begin{aligned}2t\\5t\\t\\t \end{aligned}\right] =t\left[\begin{aligned}2 \\5\\1\\1 \end{aligned}\right]+t\left[\begin{aligned}-1\\-2 \\0\\1 \end{aligned}\right].$$
(b) The nullity is $1$.
(c) The rank of $A$ is $3$.
Since $A$ has $4$ columns, one can see that
$$\text{rank}(A)+\text{nullity}(A)=3+1=4.$$
