First find a matrix $B$ that is row-equivalent to $A$ and in reduced row-echelon form
\begin{equation*}
B=
\begin{bmatrix}
\leading{1} & 0 & 3 & -2\\
0 & \leading{1} & 1 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
By Theorem BCS we can choose the columns of $A$ that have the same indices as the pivot columns ($D=\set{1,2}$) as the elements of $S$ and obtain the desired properties. So
\begin{equation*}
S=\set{\colvector{2\\-5\\1},\,\colvector{-1\\3\\1}}
\end{equation*}
We can write the column space of $A$ as the row space of the transpose (Theorem CSRST). So we row-reduce the transpose of $A$ to obtain the row-equivalent matrix $C$ in reduced row-echelon form
\begin{equation*}
C=
\begin{bmatrix}
1 & 0 & 8\\
0 & 1 & 3\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
The nonzero rows (written as columns) will be a linearly independent set that spans the row space of $\transpose{A}$, by Theorem BRS, and the zeros and ones will be at the top of the vectors,
\begin{equation*}
S=\set{\colvector{1\\0\\8},\,\colvector{0\\1\\3}}
\end{equation*}
In preparation for Theorem FS, augment $A$ with the $3\times 3$ identity matrix $I_3$ and row-reduce to obtain the extended echelon form,
\begin{equation*}
\begin{bmatrix}
1 & 0 & 3 & -2 & 0 & -\frac{1}{8} & \frac{3}{8}\\
0 & 1 & 1 & -1 & 0 & \frac{1}{8} & \frac{5}{8}\\
0 & 0 & 0 & 0 & 1 & \frac{3}{8} & -\frac{1}{8}
\end{bmatrix}
\end{equation*}
Then since the first four columns of row 3 are all zeros, we extract
\begin{equation*}
L=
\begin{bmatrix}
\leading{1} & \frac{3}{8} & -\frac{1}{8}
\end{bmatrix}
\end{equation*}
Theorem FS says that $\csp{A}=\nsp{L}$. We can then use Theorem BNS to construct the desired set $S$, based on the free variables with indices in $F=\set{2,3}$ for the homogeneous system $\homosystem{L}$, so
\begin{equation*}
S=\set{\colvector{-\frac{3}{8}\\1\\0},\,\colvector{\frac{1}{8}\\0\\1}}
\end{equation*}
Notice that the zeros and ones are at the bottom of the vectors.
This is a straightforward application of Theorem BRS. Use the row-reduced matrix $B$ from part (a), grab the nonzero rows, and write them as column vectors,
\begin{equation*}
S=\set{\colvector{1\\0\\3\\-2},\,\colvector{0\\1\\1\\-1}}
\end{equation*}