(a)\quad By Theorem BCS we can row-reduce $A$, identify pivot columns with the set $D$, and “keep” those columns of $A$ and we will have a set with the desired properties.
\begin{align*}
A\rref
\begin{bmatrix}
\leading{1} & 0 & -13 & -19 \\
0 & \leading{1} & 8 & 11 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
So we have the set of pivot columns $D=\set{1,\,2}$ and we “keep” the first two columns of $A$,
\begin{align*}
\set{
\colvector{3\\1\\-3},\,
\colvector{5\\2\\-4}
}
\end{align*}
(b)\quad We can view the column space as the row space of the transpose (Theorem CSRST). We can get a basis of the row space of a matrix quickly by bringing the matrix to reduced row-echelon form and keeping the nonzero rows as column vectors (Theorem BRS). Here goes.
\begin{align*}
\transpose{A}\rref
\begin{bmatrix}
\leading{1} & 0 & -2 \\
0 & \leading{1} & 3 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}
Taking the nonzero rows and tilting them up as columns gives us
\begin{align*}
\set{
\colvector{1\\0\\-2},\,
\colvector{0\\1\\3}
}
\end{align*}
An approach based on the matrix $L$ from extended echelon form (Definition EEF) and Theorem FS will work as well as an alternative approach.