You were doing well up until the line $H'^{-1} = H'^T$. There’s no particular reason to believe that $H$ is orthogonal; quite the opposite, in fact. (I’m going to switch here to the more conventional name $\mathtt P$ for the projection matrix and also make use of the block decomposition $\mathtt P = \left[\mathtt M \mid \mathbf p_4\right]$.) The first two columns of $\mathtt P$, $\mathbf p_1$ and $\mathbf p_2$, are the vanishing points of the world $x$- and $y$ axes. Unless the camera happens to be positioned just right relative to the $x$-$y$ plane, these vectors will not be orthogonal in the image. The upshot is that you have to compute the inverse of $\begin{bmatrix}\mathbf p_1 & \mathbf p_2 & \mathbf p_4\end{bmatrix}$, not its transpose.
This leads to the next potential problem: this matrix might not be invertible. The submatrix $\mathtt M$ is effectively the composition of a rotation and nonsingular affine transformation, so we know the first three columns are linearly independent, but it’s entirely possible that $\mathbf p_4 = \lambda \mathbf p_1 + \mu \mathbf p_2$, that is, that the image of the world origin lies on the vanishing line of the $x$-$y$ plane. This occurs when the camera center lies on this plane. In the specific construction in your question, this isn’t really an issue since you’re reprojecting onto the $x$-$y$ plane, but keep this in mind when generalizing to other planes.
Thus, the reprojection onto the $x$-$y$ plane is given by the homography matrix $$\mathtt H = \begin{bmatrix}\mathbf p_1 & \mathbf p_2 & \mathbf p_4\end{bmatrix}^{-1}.$$ This can be generalized to any plane that doesn’t contain the camera center by inserting an appropriate coordinate transformation $\mathtt B$ into the cascade, i.e., start with $$w\begin{bmatrix}x_c \\ y_c \\ 1 \end{bmatrix} = \mathtt{PB}\begin{bmatrix}x_w\\y_w\\0\\1\end{bmatrix}.$$
For comparison, here’s a construction that makes direct use of back-mapping the image points. Given a plane with homogeneous coordinate vector $\mathbf\Pi$, its intersection with the line through a fixed point $\mathbf C$ not on the plane and an arbitrary point $\mathbf X$ is $\left(\mathbf\Pi^T\mathbf X\right) \mathbf C - \left(\mathbf C^T\mathbf\Pi\right) \mathbf X$. The world coordinates of the camera center can be recovered from $\mathbf P$: its inhomogeneous Cartesian coordinates are $\tilde{\mathbf C}=\mathtt M^{-1}\mathbf p_4$. Finally, an image point $\mathbf x$ back-projects to a ray that intersects the plane at infinity at $\left((\mathtt M^{-1}\mathbf x)^T,0\right)^T$. Putting this all together, and adding a final matrix $\mathtt B$ that imposes a coordinate system on the plane, we get $$\mathtt H = \mathtt B \left(\mathbf C \mathbf \Pi^T-\mathbf C^T \mathbf \Pi \mathtt I_4 \right) \begin{bmatrix}\mathtt M^{-1}\\\mathbf 0^T\end{bmatrix}.$$ With $\mathbf\Pi=(0,0,1,0)^T$—the $x$-$y$ plane—and $$\mathtt B = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{bmatrix}$$ this produces the same homography matrix as above.
The details of the anatomy of the projection matrix that I used above can be found in Hartley and Zisserman’s Multiple View Geometry In Computer Vision and other standard references on the subject.
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
Projections are the subset of transformations which reduce the size of a space. That is, all projections are transformations, but not all transformations are projections.
Projections often come up in linear algebra and functions between real numbers, in which a projection means that the dimensionality of the space is reduced. As that seems to be the case here, let's say that $x_w, y_w, x_v, y_v \in \mathbb{R}$. Then $(x_w, y_w) \rightarrow (x_v, y_v)$ is a transformation, while $(x_w, y_w) \rightarrow x_v$ would be both a transformation and a projection.