Add a $4\times 4$ identity matrix to the right of $A$ to form the matrix $M$ and then row-reduce to the matrix $N$,
\begin{equation*}
M=
\begin{bmatrix}
-5 & 3 & -1 & 1 & 0 & 0 & 0 \\
-1 & 1 & 1 & 0 & 1 & 0 & 0 \\
-8 & 5 & -1 & 0 & 0 & 1 & 0 \\
3 & -2 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 2 & 0 & 0 & -2 & -5 \\
0 & \leading{1} & 3 & 0 & 0 & -3 & -8 \\
0 & 0 & 0 & \leading{1} & 0 & -1 & -1 \\
0 & 0 & 0 & 0 & \leading{1} & 1 & 3
\end{bmatrix}
=N
\end{equation*}
To apply Theorem FS in each of these four parts, we need the two matrices,
\begin{align*}
C&=
\begin{bmatrix}
\leading{1} & 0 & 2 \\
0 & \leading{1} & 3
\end{bmatrix}
&
L&=
\begin{bmatrix}
\leading{1} & 0 & -1 & -1 \\
0 & \leading{1} & 1 & 3
\end{bmatrix}
\end{align*}
(a)
\begin{align*}
\rsp{A}
&=\rsp{C}&&\knowl{./knowls/theorem.FS.knowl}{\text{Theorem FS}}\\
&=\spn{\set{\colvector{1\\0\\2},\,\colvector{0\\1\\3}}}&&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}}
\end{align*}
(b)
\begin{align*}
\csp{A}
&=\nsp{L}&&\knowl{./knowls/theorem.FS.knowl}{\text{Theorem FS}}\\
&=\spn{\set{\colvector{1\\-1\\1\\0},\,\colvector{1\\-3\\0\\1}}}&&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}
\end{align*}
(c)
\begin{align*}
\nsp{A}
&=\nsp{C}&&\knowl{./knowls/theorem.FS.knowl}{\text{Theorem FS}}\\
&=\spn{\set{\colvector{-2\\-3\\1}}}&&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}
\end{align*}
(d)
\begin{align*}
\lns{A}
&=\rsp{L}&&\knowl{./knowls/theorem.FS.knowl}{\text{Theorem FS}}\\
&=\spn{\set{\colvector{1\\0\\-1\\-1},\,\colvector{0\\1\\1\\3}}}&&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}}
\end{align*}