Your strategy looks good. But the computation you carried out may have some errors. I did the lengthy elementary row operations for the augmented matrix in MAPLE to get
$$\left[ {\begin{array}{*{20}{c}}
{4 - \lambda }&{ - 2}&{ - 1}&1\\
0&\lambda &{ - \frac{{2\left( { - 3 + \lambda } \right)}}{{ - 5 + \lambda }}}&{ - 2}\\
0&0&{\frac{{{\lambda ^3} - 9{\lambda ^2} + 15\lambda + 9}}{{ - 5 + \lambda }}}&{ - \lambda - 3}
\end{array}} \right]$$
while during the process of elimination it is assumed that $\lambda\ne 0,4,5$. Now, if you look at the third row which is an equation you get
$$\left( {\frac{{{\lambda ^3} - 9{\lambda ^2} + 15\lambda + 9}}{{ - 5 + \lambda }}} \right){x_3} = - \lambda - 3$$
The system can have no solutions only when this equation can lead to a contradiction. This can happen when the coefficient of $x_3$ in LHS is zero and the RHS is nonzero. If this happens we can conclude that the system doesn't have a solution for specified values of $\lambda$. So we may write
$\left\{ \begin{array}{l}
{\lambda ^3} - 9{\lambda ^2} + 15\lambda + 9 = 0\\
- \lambda - 3 \ne 0
\end{array} \right.\,\,\,\,\, \to \,\,\,\,\,\left\{ \begin{array}{l}
\lambda = 3,3 \pm 2\sqrt 3 \\
\lambda \ne - 3
\end{array} \right.\,\,\,\, \to \,\,\,\,\lambda = 3,3 \pm 2\sqrt 3 $
Since this is consistent with our previous assumptions that $\lambda\ne 0,4,5$, we can say that for these values of $\lambda$ the system doesn't have a solution. Consequently, your second statement is false since we proved that the system has no solutions for some values of $\lambda$. Also, as we have found some values for $\lambda\ne -1,5$ that the system does not have a solution, your first statement is false too.
For (a):
If $a = b = 0$ then adding $2$ times the second equation from the first gives $x = 0$. The solution space is then $\{(x,y,z)\in\mathbb{R}^3:x = 0\}$ (can you justify this statement?), which is a two-dimensional subspace of $\mathbb{R}^3$.
If we don't have $a=b=0$, this means either $a=b\neq 0$, or $a\neq b$. In the first case, the same calculation as above gives you $x = a+2b$. But for any $C\neq 0$, the set $$\{(x,y,z)\in\mathbb{R}^3:x=C\}$$ is not a subspace, since it does not contain the point $(0,0,0)$. So the solution space does not form a subspace.
On the other hand, if $a\neq b$, then the same calculation gives $x = a+2b$ again. In this case, either $x = 0$, in which case substituting this into the original set of equations gives something inconsistent (why?), or $x\neq 0$, in which case the solution space is not a subspace (why?).
For (b):
You now know that the solution space is the subspace $\{(x,y,z)\in\mathbb{R}^3:x = 0\}$. What are two vectors that span this space that are not multiples of each other?
Best Answer
Problem
$$ \mathbf{A} = \left[ \begin{array}{rrr} 0 & 1 & -2 \\ -1 & 2 & -1 \\ 2 & -4 & 3 \\ 1 & -3 & 2 \\ \end{array} \right] $$
Associated Permutation Matrix
Don't start with a $0$ pivot element. Move the first row down. The permutation matrix interchanges the first two rows.
$$ \mathbf{P} = \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right] $$
Input
$$ \begin{align} \mathbf{P} \mathbf{A} = % P \left[ \begin{array}{cccc} 0 & \boxed{1} & 0 & 0 \\ \boxed{1} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right] % A \left[ \begin{array}{rrr} 0 & 1 & -2 \\ -1 & 2 & -1 \\ 2 & -4 & 3 \\ 1 & -3 & 2 \\ \end{array} \right] % &= % L \left[ \begin{array}{rrr} -1 & 2 & -1 \\ 0 & 1 & -2 \\ 2 & -4 & 3 \\ 1 & -3 & 2 \\ \end{array} \right] \end{align} $$
Decomposition
$$ \begin{align} \mathbf{P} \mathbf{A} &= \mathbf{L} \mathbf{U} \\ % PA \underbrace{\left[ \begin{array}{rrr} -1 & 2 & -1 \\ 0 & 1 & -2 \\ 2 & -4 & 3 \\ 1 & -3 & 2 \\ \end{array} \right]}_{\color{blue}{m}\times \color{red}{n}} % &= % L \underbrace{\left[ \begin{array}{rrrc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -2 & 0 & 1 & 0 \\ -1 & -1 & -1 & 1 \\ \end{array} \right]}_{\color{blue}{m}\times \color{blue}{m}} % U \underbrace{\left[ \begin{array}{rrr} -1 & 2 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right]}_{\color{blue}{m}\times \color{red}{n}} \end{align} $$