I want to see if my work is justifiable. I am tasked with the following: 
I will neglect to prove (a), as the work for this is fairly straight forward. I will center my attention on (b).
$$\text{Implication is} \ \mathscr {B}_{\mathbb R^4 / \ ker \ T} = \left\{ \begin{bmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} \end{bmatrix},\begin{bmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} \end{bmatrix} \right\} \text{, where $\mathscr B$ is the basis for $\mathbb R^4 / \ ker \ T$}$$
$$ \implies \text{for some $[v] \in \mathbb R^4 / \ ker \ T$,}$$ $$[v] = a_1 \begin{bmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} \end{bmatrix} + a_2 \begin{bmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} \end{bmatrix}$$
$$\implies \begin{bmatrix} \begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix} \end{bmatrix} = a_1 \begin{bmatrix} \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} \end{bmatrix} + a_2 \begin{bmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} \end{bmatrix}$$
$$\implies \begin{bmatrix} \begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix} \end{bmatrix} = \begin{bmatrix} \begin{pmatrix} 0 \\ a_1 \\ a_2 \\ 0 \\ \end{pmatrix} \end{bmatrix} $$
$$\implies \begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix} - \begin{pmatrix} 0 \\ a_1 \\ a_2 \\ 0 \\ \end{pmatrix} \in ker \ T$$ $$ \implies \begin{pmatrix} x \\ y-a_1 \\ z-a_2 \\ t \\ \end{pmatrix} = b_1 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} + b_2 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ \end{pmatrix} \text{, where $b_1, b_2 \in ker \ T$}$$ After some algebra...
$$\implies a_1 = y - x$$ $$\implies a_2 = z - t$$
Which I think implies $\mathscr B$ is spanning? $x,y,z,t$ are all arbitrary. Now I am left with proving $\mathscr B$ is linearly independent. I may skip a step or two.
$$ \begin{bmatrix} \begin{pmatrix} 0 \\ a_1 \\ a_2 \\ 0 \\ \end{pmatrix} \end{bmatrix} = [0] \iff a_1 = a_2 = 0 $$
$$\begin{bmatrix} \begin{pmatrix} 0 \\ a_1 \\ a_2 \\ 0 \\ \end{pmatrix} \end{bmatrix} = [0] \implies \begin{pmatrix} 0 \\ a_1 \\ a_2 \\ 0 \\ \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} \in Ker \ T$$
$$\implies \begin{pmatrix} 0 \\ a_1 \\ a_2 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} b1 \\ b1 \\ b2 \\ b2 \\ \end{pmatrix} \implies a_1 = a_2 = 0$$
Hence, basis. Has what I done made sense?
It's all correct.
Note that the 4 given vectors form a basis of $\Bbb R^4$, two of them spans the kernel, so the (images of) the other two will be a basis of the quotient space.