We claim that $\{AB - BA : A,B \in M_n(\mathbb{C})\} = \{A \in M_n(\mathbb{C}), \operatorname{Tr }A = 0\} = \ker \operatorname{Tr}$.
We already know that $\{AB - BA : A,B \in M_n(\mathbb{C})\} \subseteq \ker \operatorname{Tr}$.
Let $E_{ij}$ denote the matrix with $1$ at the position $(i,j)$ and $0$ elsewhere.
Check that $B = \{E_{ij} : 1 \le i, j \le n, i\ne j\} \cup \{E_{ii} - E_{nn} : 1 \le i \le n-1 \}$ is a basis for $\ker \operatorname{Tr}$.
For $1 \le i, j \le n, i\ne j$ we have
$$E_{ij} = E_{ik}E_{kj} - E_{kj}E_{ik}$$
where $k$ is some index $\ne i,j$. To see this, let $\{e_1, \ldots, e_n\}$ be the standard basis for $\mathbb{C}^n$ and note that $E_{ij}e_j = e_i$ and $E_{ij}e_r = 0$ for $r \ne j$. Now verify that
$$(E_{ik}E_{kj} - E_{kj}E_{ik})e_r =
\begin{cases}
0, &\text{if } r \ne j,k\\
E_{ik}E_{kj}e_j = E_{ik}e_k = e_i, &\text{if }r = j\\
-E_{kj}E_{ik}e_k = -E_{kj}e_i = 0, &\text{if }r = k\\
\end{cases}$$
For $1 \le i \le n-1$ we have
$$E_{ii} - E_{nn} = E_{in}E_{ni} - E_{ni}E_{in}$$
Indeed
$$(E_{in}E_{ni} - E_{ni}E_{in})e_r =
\begin{cases}
0, &\text{if } r \ne i,n\\
E_{in}E_{ni}e_i = E_{in}e_n = e_i, &\text{if }r = i\\
- E_{ni}E_{in}e_n = -E_{ni}e_i = -e_n, &\text{if }r = n\\
\end{cases}$$
Therefore $B \subseteq \{AB - BA : A,B \in M_n(\mathbb{C})\}$ so we conclude $\ker \operatorname{Tr} \subseteq \{AB - BA : A,B \in M_n(\mathbb{C})\}$.
a) The image is $\{0\}$, not $0$. And the rank is $0(=\dim\{0\})$. You are right about the null space and the nullity.
b) The image is $V$, not $v$. And, although indeed the nullity is $\dim_F\{0\}$, I would have said that it is $0$.
c) The image if $F\times\{0\}$, and the rank is $1$. The rest is fine.
Best Answer
As $f$ is a non zero linear functional, there exists $x_0 \in V$, such that $$f(x_0)=\alpha\implies \frac{1}{\alpha}f(x_0)=f\left(\frac{1}{\alpha}x_0\right)=1.$$ for some $\alpha \neq 0$.
Then for any $\beta \in \mathbb{F}$, $$\beta=\beta .1=\beta f\left(\frac{1}{\alpha}x_0\right)=f\left(\frac{\beta}{\alpha}x_0\right)$$
Hence every element of the $\mathbb{F}$ could be hit as a scalar multiple of $x_0$ and and $1 \in \mathbb{F}$ generates $\mathbb{F}$ , hence rank of $f$ is 1.