Assume that A is row equivalent to B. Find bases for Nul A and Col A.

Since B is a row echelon form of A, we see that the first, third and fifth columns of A are its pivot columns. Thus a basis fo Col A is
$\{\left[\begin{array}{c}1\\ 2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-5\\ -5\\ 0\\ -5\end{array}\right],\left[\begin{array}{c}-3\\ 2\\ 5\\ -2\end{array}\right]\}$
To find a basis for Nul A, we find the general solution of Ax=0 in terms of the free variables. Since it is row equivalent to B we can simply get reduced row echelon form of B:
$\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]\begin{array}{c}{R}_2\to \frac{1}{5}{R}_2\\ R_3\to \frac{1}{-9}{R}_3\\ \\ \end{array}\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& \frac{-7}{5}& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]$
$\begin{array}{c}{R}_1\to R_1-5R_3\\ R_2\to R_2-\frac{8}{5}{R}_3\end{array}\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 1& \frac{-7}{5}& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]$
to get: ${\displaystyle x_1=-2x_2-4x_4,x_3=\frac{7}{5}{x}_4,x_5=0}$, with ${\displaystyle x_2}$ and ${\displaystyle x_4}$ free. So
$x=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\end{array}\right]=x_2\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right]+x_4\left[\begin{array}{c}-4\\ 0\\ 7/5\\ 1\\ 0\end{array}\right]$
And a basis for Nul A is
$\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 7/5\\ 1\\ 0\end{array}\right]\}$
Result: Basis for Col A is:
$\{\left[\begin{array}{c}1\\ 2\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}-5\\ -5\\ 0\\ -5\end{array}\right],\left[\begin{array}{c}-3\\ 2\\ 5\\ -2\end{array}\right]\}$
Basis for Nul A is
$\{\left[\begin{array}{c}-2\\ 1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}-4\\ 0\\ 7/5\\ 1\\ 0\end{array}\right]\}$