Answer
a basis for the nullspace of $A$ consists of the vectors
$$\left[\begin{aligned}2\\ -1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} -5\\ 1\\0\\1 \end{aligned}\right].$$
Work Step by Step
Given the matrix
$$
\left[ \begin {array}{cccc} 1&4&2&1\\ 0&1&1&-1
\\ -2&-8&-4&-2\end {array} \right]
.
$$
The reduced row echelon form is given by
$$
\left[ \begin {array}{cccc} 1&0&-2&5\\ 0&1&1&-1
\\ 0&0&0&0\end {array} \right]
.
$$
The corresponding system is
$$
\begin{aligned} x_{1} -2x_3+5x_4&=0
\\
x_{2} +x_3-x_4&=0
\end{aligned}.
$$
The solution of the above system is $x_1=2s-5t$,$x_2=-s+t$, $x_3=s$ and $x_4=t$. This means that the solution space of $Ax = 0 $ consists of the vectors on the following form
$$x= \left[\begin{aligned} x_{1}\\ x_{2} \\x_3 \\x_4 \end{aligned}\right]= \left[\begin{aligned} 2s-5t\\-s+t\\s\\t \end{aligned}\right]=s\left[\begin{aligned}2\\ -1\\1 \\0 \end{aligned}\right]+t\left[\begin{aligned} -5\\ 1\\0\\1 \end{aligned}\right] .$$
So, a basis for the nullspace of $A$ consists of the vectors
$$\left[\begin{aligned}2\\ -1\\1 \\0 \end{aligned}\right], \left[\begin{aligned} -5\\ 1\\0\\1 \end{aligned}\right].$$
