You have the correct rank of $A$, indeed $\text{rank}(A) = 2$. First, you have the row-reduced echelon form incorrect...it should be $$\begin{pmatrix} 1 & 0 & -5 & 1 \\ 0 & 1 & -3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \,.$$
In regards to your question about multiplying rows by scalars in the process of row reduction, this is perfectly fine. Writing the above in parametric form, we have
$$\begin{cases} x_1 &= \,\,\,\, 5x_3-x_4 \\ x_2 &= \,\,\,\,3x_3 -4x_4 \end{cases}$$
with $x_3,x_4$ free. This is equivalent to the vector solution
$$\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 5 \\3 \\1\\0 \end{pmatrix}x_3 + \begin{pmatrix} -1 \\ -4 \\0\\1 \end{pmatrix}x_4 \,\,.$$
Thus, the basis for the null space that you were seeking is
$$ \mathscr{B} =\left\{ \begin{pmatrix} 5 \\3 \\1\\0 \end{pmatrix},\begin{pmatrix} -1 \\ -4 \\0\\1 \end{pmatrix} \right\} \,.$$
Remember, you have no solution to a system if say, you have a system
$$
\left[
\begin{array}{ccc c|c}
1 & 0 & 0 & 0 & a \\
0 & 1 & 0 & 0 & b \\
0 & 0 & 0 & 0 & c\\
\end{array}
\right]$$
where $c\neq 0$. This is because the system essentially says $0=c$.
The nullspace is the subspace of all solutions to $A\mathbf{x}=\mathbf{0}$, and the column space is the span of the columns.
To find the null space, you simply want to determine a basis for the solution vectors of the homogeneous system. For the reduced row-echelon form of your homogeneous system, you have
$$
\left[
\begin{array}{ccc c|c}
1 & 0 & 0 & 2 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0\\
\end{array}
\right]$$
Let's say each respective column corresponds to the variables $x_{1}$, $x_{2}$, $x_{3}$, and $x_{4}$. Since the column for $x_{4}$ does not have a leading $1$, we let $x_{4}$ be the free variable.
Then the RREF gives us the following solution to the homogeneous system:
$$x_{1}+2x_{4}=0\ \mathrm{or}\ x_{1}=-2x_{4}$$
$$x_{2} =0$$
$$x_{3}+x_{4}=0\ \mathrm{or}\ x_{3}=-x_{4}$$
In vector form, we have
$$\begin{bmatrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{bmatrix}=\begin{bmatrix}-2x_{4} \\ 0 \\ -x_{4} \\ x_{4}\end{bmatrix}=\begin{bmatrix}-2x_{4} \\ 0x_{4} \\ -x_{4} \\ x_{4}\end{bmatrix}=\begin{bmatrix}-2 \\ 0 \\ -1 \\ 1\end{bmatrix}x_{4}.$$
So the nullspace of your original matrix is the span of the vector $\begin{bmatrix}-2 \\ 0 \\ -1 \\ 1\end{bmatrix}$, since $x_{4}$ can take any value.
For the column space, you need to look at the columns in the RREF that have leading $1$'s. the column space will be the span of the columns from your original matrix which have a leading $1$ in the RREF (i.e. the column space is the span of columns $1$, $2$, and $3$).
You should post the original matrix that you're working with if you are going to ask this type of question, so that we can fully answer your question and ensure you did not make any mistakes after applying row operations.
Best Answer
Write the system
$$\left\{\begin{array}{rcl} x-2y+2z+4t& =& 0 \\ z+t & = & 0\end{array}\right.$$
as
$$\left\{\begin{array}{rcl} x+2z& =& 2y-4t \\ z & = & -t\end{array}\right.$$ and solve it. You get,
$$\left\{\begin{array}{rcl} x& =& 2y-2t \\ z & = & -t\end{array}\right.$$ That is, $$(2y-2t,y,-t,t)$$ is an element of the null space for any $y,t.$ Now, look for two linearly independent vectors.
(Note that the kernel has dimension $2.$ So the system has infinitely many solutions that have to depend on two parameters.)