Your question is not well clear, but I'm attempting an answer.
First note that the usual definition is that
a linear transformation $P$ is a projector if $P^2=P$
Your additional condition $P^T=P$ caracterize the orthogonal projectors (in a real vector space $V$), where:
$P$ is orthogonal if the kernel of $P$ is orthogonal to its range.
Now note that for any vector $y \in V$ we have that $(y-Py)$ is an element of $\ker(P)$ because $P(y-Py)=Py -P^2y=0$.
So, $P$ is an orthogonal projector if $\forall x,y \in V$ we have :
$$
\langle Px,(y-Py) \rangle=0
$$
But this is done because:
$$
\langle Px,(y-Py) \rangle=\langle P^2x,(y-Py) \rangle=\langle Px,P^T(y-Py) \rangle=\langle Px,(Py-P^2y) \rangle=\langle Px,0 \rangle=0
$$
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
If $P$ is an orthogonal projection, then it follows that $P = P^T$ and hence $N(P) = N(P^T)$. However, if $P$ is not orthogonal, then it may not be true that $N(P) = N(P^T)$. For example, consider the (non-orthogonal $\iff$ non-symmetric) projection matrix
$$ P = \left[ \begin{array}{cc} 0 & 1\\ 0 & 1\\ \end{array} \right] $$
It is clear that $N(P)=\{(a,0)\in\mathbb{R}^2\,|\, a\in\mathbb{R}\}$ and $N(P^T) = \{(a,-a)\in\mathbb{R}^2\,|\, a\in\mathbb{R}\}$. The only point they have in common is the zero vector.