Idea : Create a new vector space in which $T$ is diagonalizable i.e. has 1D invariant subspaces. Pull back all the subspaces to $V$ via an appropriate map, wherein all the invariant subspaces get mapped back to the desired disjoint subspaces.
We do the following : I claim that $V$ can be given a vector space structure over the complex numbers (or a field isomorphic to the complex numbers).
More precisely, consider $\mathbb Q(T)$, the field created by taking all rational functions in $T$ with real coefficients (so elements of this field are matrices of the form $p(T)(q(T))^{-1}$ where $p,q$ are polynomials with real coefficients). One easily sees that this field is isomorphic to $\mathbb Q(i)$, by sending $x+iy \to xI + yT$ and noting the relation satisfied by $T$.
To make $V$ a vector space over $\mathbb Q(T)$ is obvious : addition is as usual and scalar multiplication is defined by $(aI+bT)v = av + b(Tv)$.
Denote $V_{\mathbb Q(T)}$ for $V$ with changed base field.
Once $V$ is an $\mathbb Q(T)$ subspace, the operator $T : V_{\mathbb Q(T)} \to V_{\mathbb Q(T)}$ (now scalar multiplication by the field element $T$) is obviously a diagonal operator regardless of basis, a multiple of the identity matrix. Consequently, we get that $V_{\mathbb Q(T)}$ breaks into $1$ dimensional subspaces which are invariant under $T$ (obviously, taking any basis, the elements of this basis are all eigenvectors, so form one dimensional invariant subspaces). This is similar to how scalar multiplication (in the usual setting) does not change matrix regardless of change of basis.
We now perform a restriction of scalars as follows. Define a map $i : V_{\mathbb R} \to V_{\mathbb Q(T)}$, by the identity map (!)
I claim that any one dimensional subspace $\overline{\{b\}}$ of $V_{\mathbb Q(T)}$ has a two dimensional preimage. But this is obvious : $\overline{\{b\}}$ in $\mathbb Q(T)$ consists of all the elements $(xI+yT)b = xb + yTb \in \overline{\{b,Tb\}_{\mathbb Q}}$, and $b,Tb$ are linearly independent over $\mathbb Q$. Furthermore this is $T$ invariant. (All easy checks)
Thus, if $b_1,...,b_n$ is a basis of $V_{\mathbb Q(T)}$, then $i^{-1} (\overline{\{b_j\}})$ is a two dimensional $T$ invariant subspace of $V_{\mathbb Q}$. Finally, these subspaces are disjoint, because their images under $i$ lie in different subspaces of $V_{\mathbb Q(T)}$.
It follows that $V = \oplus i^{-1}(b_j)$.
Best Answer
Let's consider the kernel and image of $T$. The kernel of $T$ must be two dimensional, since it cannot be three dimensional since $T\ne 0$, and if it were one dimensional, then we would have that the rank of $T$ would be two by rank-nullity, so we couldn't have $T^2=0$. Hence the dimension of the image of $T$ must be one dimensional. Also it is clear that the image of $T$ is a one dimensional subspace of $\ker T$.
Now, if $V\subseteq F^3$ is $T$-invariant, then $T(V)\subseteq V$. Since the image of $T$ is one dimensional, either $V$ contains the image of $T$, or $V$ lies in the kernel of $T$.
If $V$ is two dimensional, then if it lies in the kernel of $T$, then it is the kernel of $T$, and hence contains the image of $T$. Thus the two dimensional invariant subspaces are precisely those containing the image of $T$. To count the number of planes containing a line in $F^3$, fix a complement of the line. Then planes containing the line in $F^3$ correspond to lines in that complement, of which there are $(p^2-1)/(p-1)=p+1$.
Conversely if $V$ is one dimensional, then if it contains the image of $T$, it is the image of $T$, so it lies in the kernel. Thus the one dimensional invariant subspaces for $T$ are the one dimensional subspaces of the kernel of $T$. We just counted the number of one dimensional subspaces of a two dimensional $F$ vector space. There are $p+1$ one-dimensional invariant subspaces.
I'm not sure exactly what the second question is asking us to count at all, but if you can clarify that it'd be cool. Might be because it's late, since I'm having trouble even formulating guesses as to what it might be supposed to mean, which suggests my brain is about to cease functioning.
Intuition?
Hard to say exactly what the intuition is, but I guess the idea is that since $T$ has nontrivial kernel, that is easily invariant (as is any subspace contained in it), so let's start looking into that. Similarly, since the kernel is nontrivial, $T$ is not surjective, and the image of $T$ will also be invariant (and any subspace containing it), so it's worth looking into that.