Hints:
(a) If $T$ is rank one, $T$ should be of the form $T=\sigma_1 uv^T$ where $\sigma_1$ is the highest singular value and $u$,$v$ are left and right singular vectors respectively. Convince yourself that this is true using singular value decomposition. You might need the cyclic property of trace as well.
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$$0=trace(T)=trace(\sigma_1uv^T)=\sigma_1v^Tu$$
and even more
$$T^2=\sigma_1^2uv^Tuv^T=\sigma_1^2uv^T(v^Tu)$$
(b) Use the fact that $A=aI+\sigma_1 uv^T$. Observe what happens when $T$ is rank-one and zero-trace. Now say $B=\alpha I+\beta uv^T$ for some unknown constants $\alpha$ and $\beta$. Try to find if there are any $\alpha$ and $\beta$ such that $AB=I$ and $BA=I$. If you are still struggling, take a look at sherman-morrison formula.
(c) use the same strategy as in (b).
Let $J= \operatorname{diag}(J_1,...,J_k)$ be the Jordan normal form.
Note that $J^2= \operatorname{diag}(J_1^2,...,J_k^2)$
The rank of $J$ is given by the sum of the
ranks of the blocks that is, $\operatorname{rk} J = \sum_k \operatorname{rk} J_k$.
Similarly,
$\operatorname{rk} J^2 = \sum_k \operatorname{rk} J_k^2$.
It follows from the hypothesis that the rank of the Jordan block
corresponding to the zero eigenvalue is zero. That is, the Jordan
block is identically zero (and so is the square of the Jordan block).
For the blocks $J_k$ corresponding to non zero eigenvalues, we have
$\operatorname{rk} J_k = \operatorname{rk} J_k^2$.
It follows that $\operatorname{rk} J = \operatorname{rk} J^2$.
Alternative:
Let $N = \ker A$. We see that $z=\dim N$ is the number of zero eigenvalues of $A$. Let $b_1,..,b_z$ be a basis for $N$ and complete the basis with $b_{z+1},...,b_n$. Note that $N$ is $A$ invariant and so in this basis, $A$ has the form
$\begin{bmatrix} 0 & A_{12} \\ 0 & A_{22}\end{bmatrix}$. Furthermore, we
must have $\det A_{22} \neq 0$ otherwise $A$ would have more that $ z$ zero eigenvalues.
Then we want to show that $\ker A^2 = N$. Note that $A^2$ has the form
$\begin{bmatrix} 0 & A_{12}A_{22} \\ 0 & A_{22}^2\end{bmatrix}$ and it follows from
this that if $x \in \ker A^2$ then $x \in \ker A$.
Best Answer
From $A^2=A$ you get that the only possible eigenvalues of $A$ are $0$ and $1$.
From the fact that $A^2=A$ you get that any vector $x$ can be written as $x=Ax + (I-A)x$ and that this representation is unique. In other words, the range of $A$ and the range of $I-A$ are complementary subspaces.
By the Rank-Nullity Theorem, $\operatorname{rank}A=\dim\ker (I-A)$. And $\ker (I-A)$ is precisely the eigenspace corresponding to the eigenvalue $1$. Similarly, $\operatorname{rank}(I-A)=\dim\ker A$.
If we let $e_1,\ldots,e_m$ be an ortonormal basis of $\ker A$ and $e_{m+1},\ldots,e_n$ an orthonormal basis of $\ker (I-A)$, then $$ \operatorname{Tr}(A)=\sum_k\langle Ae_k,e_k\rangle=\sum_{k=m+1}^n \langle Ae_k,e_k\rangle=\sum_{k=m+1}^n \langle e_k,e_k\rangle=\sum_{k=m+1}^n1=\operatorname{rank}(A). $$