T-invariant subspaces I'm studying for a qualifying exam, and I'd really appreciate some help on the following questions.
Let p be a prime integer, $F=\mathbb{Z/pZ}$, V a vector space over F, and $T\in\mathscr{L}(V)$. Assume $c_T=x^3$ and $m_T=x^2$.
1) Express V as a direct sum of cyclic F[x]-modules.
Solution: $V=F[x]/(x^2)\bigoplus F[x]/(x)$.
 This gives dim$V=3$. The corresponding Jordan form is 
$$\begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$ with corresponding basis $\mathscr{B}=\{a_1,xa_1,a_2\}$. $|V|=p^3$  
2) How many 1-dimensional T-invariant subspaces does V have?
Solution: The 1-dimensional T-invariant subspaces are $<c_1xa_1+c_2a_2>$ with $c_i\in F$. So, there are $p^2$ subspaces.
3) How many of the 1-dimensional T-invariant subspaces of V are direct summands of V?
Solution: 2p
4) How  many 2-dimensional T-invariant subspaces does V have?
These include subspaces $<a_1,xa_1+c_2a_2>$, of which $p$ are 2 dimensional. Also,$<a_1,a_2>$, of which there is 1.  Also, $<xa_1,a_1+c_2a_2>$ of which $p$ are 2 dimensional. We also have $<xa_1,a_2>$, which is another.   Finally, $<a_2,a_1+c_1xa_1>$. There are $p-1$ of these not counted already. So, a total of $3p+1$.
5) How many of the 2-dimensional T-invariant subspaces of V are direct summands of V?
Solutions: $(p-1)^2$
I believe most of what I have done is incorrect, and I don't really understand parts 2-5. Thanks in advance.

Part 2: Right basic idea, but you've over-counted the correct subspaces and included $c_1 = c_2 = 0$, which isn't $1$-dimensional.
Our subspaces are
$\langle xa_1 + c_2 a_2\rangle $ and $\langle a_2 \rangle$. All together, that's $p+1$ subspaces.
Part 3: Not sure what they're going for here.  Any subspace can be a direct summand of $V$ under a certain definition.  If we want the complementary summand to be $T$-invariant as well, however, only $\langle a_2 \rangle$ will do.
Part  4: same mistake as with 2.  See if you can come up with the correct answer here.
Part 5: again, depends on their definition of "direct summand".

Full answer to part 4:
We can take the direct sum of any of the two of the spaces from 2.  There are $p+1$ such one-dimensional spaces, so we can combine pairs to get $\binom{p+1}{2} = \frac 12 p(p+1)$ distinct spaces in total.
Our second option is any space of the form $\langle a_1 + c a_2, xa_1 \rangle$.  There are $p$ such spaces, corresponding to all choices of $c$.
So, all together, we have $p + \frac 12 p(p+1) = \frac 12 p(p+3)$.
