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 $4 - 2 = 2$.
A matrix $P$ is called a projection if $P^2=P$. From this, it is easy to see that all eigenvalues must be $0$ or $1$. On the other hand, a matrix with only eigenvalues $0$ and $1$ may not necessarily a projection.
Example. Take
$$A=\begin{pmatrix}
\color{red}1&0&0\\
0&\color{red}0&\color{blue}1\\
0&0&\color{red}0
\end{pmatrix}.$$
This matrix is not diagonalizable. It is already in its Jordan normal form. Its eigenvalues are $\{\color{red}1,\color{red}0,\color{red}0\}$ but
$$A^2=\begin{pmatrix}
\color{red}1&0&0\\
0&\color{red}0&\color{blue}0\\
0&0&\color{red}0
\end{pmatrix}\not=A.$$
Any projection onto a line (through the origin) spanned by a normalized vector $a\in\Bbb R^3$ can be given as
$$x\quad\mapsto\quad a\cdot\langle x,\;a\rangle_M$$
for some inner product $\langle x,\;a\rangle_M:=x^\top M a$ (where $M$ is a symmetric positive definite matrix). So there are more such projections like the one you have given, but the only difference is that the inner product can be a general one and must not be the standard product of $\Bbb R^3$.
Such projections with $M\neq I$ (where $I$ is the identity matrix) might not be "perpendicular" as you know them from a geometric point of view. But they still obey the definition of a projection.
Best Answer
The rank of a matrix $A$ is, by definition, the dimension of its column space, that is, the dimension of the range of the associated linear map $\varphi_A:=x\mapsto Ax$.
If $A$ (or rather, $\varphi_A$) is a (not necessarily orthogonal) projection to a subspace $W$, then its range is simply $W$ and thus, its rank is indeed the same as $\dim W$ (and, btw, its kernel determines the 'direction' of the projection).