Let $R$ be the row equivalent form of the matrix $A$. Then we know that the solution of $Ax=0$ is same as the solution of $Rx=0$. Note that $R$ is any row-equivalent form of $A$, need not be an row-reduced equivalent form.
By applying elementary row operation on the given matrix $A$ we get
$$\left[\begin{array}{cccc}1 & 3 & 0 & -4 \\ 2 & 6 & 0 & -8\end{array}\right] \xrightarrow{R_2\leftarrow R_2-2R_1} \left[\begin{array}{cccc}1 & 3 & 0 & -4 \\ 0 & 0 & 0 & 0\end{array}\right]$$
By taking $R= \left[\begin{array}{cccc}1 & 3 & 0 & -4 \\ 0 & 0 & 0 & 0\end{array}\right]$ we note that the system $Rx=0$ can be written as
$$ \left[\begin{array}{cccc}1 & 3 & 0 & -4 \\ 0 & 0 & 0 & 0\end{array}\right] \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$$
It follows that
$$x_1+3x_2-4x_4=0.$$
We have only one equation with four variable. Therefore, we have to consider three free variables. Take $x_2=k$, $x_3=l$, and $x_4=m$, where $k,l,m$ are real numbers. Then solutions of the system $Ax=0$ are of the following form
$$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}= \begin{bmatrix}4m-3k\\k\\l\\m\end{bmatrix} = \begin{bmatrix}-3\\1\\0\\o\end{bmatrix}k + \begin{bmatrix}0\\0\\1\\0\end{bmatrix} l + \begin{bmatrix}4\\0\\0\\1\end{bmatrix} m,$$
where $k,l,m$ are some real numbers.