Let $A=[v_1\;v_2\;v_3\;v_4]$ be the required matrix. You need that
$$
A\begin{bmatrix}1\\2\\0\\0\end{bmatrix}=v_1+2v_2=0
$$
so $v_1=-2v_2$; also
$$
A\begin{bmatrix}0\\1\\2\\0\end{bmatrix}=v_2+2v_3=0
$$
so $v_2=-2v_3$ and $v_1=4v_3$.
You see that $v_4$ can be anything; now just take
$$
v_3=\begin{bmatrix}1\\2\\0\\0\end{bmatrix}
\quad
v_4=\begin{bmatrix}0\\1\\2\\0\end{bmatrix}
$$
Then the matrix
$$
A=[v_1\;v_2\;v_3\;v_4]=
\begin{bmatrix}
4 & -2 & 1 & 0 \\
8 & -4 & 2 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
will have the required column space and null space: indeed, the column space is generated by $v_3$ and $v_4$, so it has dimension $2$. Since the null space contains the required one, it will be equal to it by the rank-nullity theorem.
Part (a): By definition, the null space of the matrix $[L]$ is the space of all vectors that are sent to zero when multiplied by $[L]$. Equivalently, the null space is the set of all vectors that are sent to zero when the transformation $L$ is applied. $L$ transforms all vectors in its null space to the zero vector, no matter what transformation $L$ happens to be.
Note that in this case, our nullspace will be $V^\perp$, the orthogonal complement to $V$. Can you see why this is the case geometrically?
Part (b): In terms of transformations, the column space $L$ is the range or image of the transformation in question. In other words, the column space is the space of all possible outputs from the transformation. In our case, projecting onto $V$ will always produce a vector from $V$ and conversely, every vector in $V$ is the projection of some vector onto $V$. We conclude, then, that the column space of $[L]$ will be the entirety of the subspace $V$.
Now, what happens if we take a vector from $V$ and apply $L$ (our projection onto $V$)? Well, since the vector is in $V$, it's "already projected"; flattening it onto $V$ doesn't change it. So, for any $x$ in $V$ (which is our column space), we will find that $L(x) = x$.
Part (c): The rank is the dimension of the column space. In this case, our column space is $V$. What's it's dimension? Well, it's the span of two linearly independent vectors, so $V$ is 2-dimensional. So, the rank of $[L]$ is $2$.
We know that the nullity is $V^\perp$. Since $V$ has dimension $2$ in the $4$-dimensional $\Bbb R^4$, $V^\perp$ will have dimension $4 - 2 = 2$. So, the nullity of $[L]$ is $2$.
Alternatively, it was enough to know the rank: the rank-nullity theorem tells us that since the dimension of the overall (starting) space is $4$ and the rank is $2$, the nullity must be $4 - 2 = 2$.
Best Answer
Here’s a way to think about this:
Recall that matrix multiplication corresponds to composition of linear transformations. That is, you can look at the product $ABx$ as feeding the vector $x$ first to the linear transformation represented by $B$, and then feeding the result of that to the linear transformation represented by $A$.
The null space of a matrix is the set of vectors that its associated transformation maps to zero. The only possible image of the zero vector under a linear transformation is the zero vector of the codomain. So, if $B$ maps some vector $x$ to $0$, then $A$ can’t “unmap” that to something non-zero: once a vector gets sent to zero, it stays there. This means that the null space can’t get any smaller when you add another matrix to the chain of operations. (When I say “larger” and “smaller,” I mean dimension, not cardinality.)
A similar line of reasoning can be applied to the column space of a product. The rank of a matrix—the dimension of its column space—gives you a maximum dimension for the image of a vector space under the associated linear transformation. In addition, you should be able to convince yourself that the dimension of that image can’t exceed the dimension of the space being mapped. So, the dimension of the column space can only stay the same or shrink at each stage.