If the ground field $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$, the following gives an elementary proof. Clearly every commutator has zero trace, so it suffices to show that every real or complex matrix with zero trace is a commutator. First, every traceless matrix is $\mathbb{F}$-similar to a matrix with zero diagonal (we will prove this claim later). So, WLOG we may assume that $C$ has a zero diagonal. Take $A$ as an arbitrary diagonal matrix $\mathrm{diag}(a_1,\ldots,a_n)$ with real and distinct diagonal entries. The equation $C=AB-BA$ then boils down to $(a_i-a_j)b_{ij}=c_{ij}$, which is solvable as $b_{ij}=c_{ij}/(a_i-a_j)$. QED
We now prove our claim that any traceless matrix $C$ is $\mathbb{F}$-similar to a matrix with zero diagonal when $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. In the real case, as $C$ is traceless, if it has some nonzero diagonal entries, some two of them must have different signs. WLOG assume that they are $c_{11}$ and $c_{22}$. Note that
$$
\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}
\begin{pmatrix}c_{11}&c_{12}\\c_{21}&c_{22}\end{pmatrix}
\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}
=\begin{pmatrix}c_{11}\cos^2\theta-(c_{12}+c_{21})\sin\theta\cos\theta+c_{22}\sin^2\theta&\ \ast\\ \ast&\ \ast\end{pmatrix}.
$$
Since $c_{11}$ and $c_{22}$ have different signs, $c_{11}\cos^2\theta-(c_{12}+c_{21})\sin\theta\cos\theta+c_{22}\sin^2\theta=0$ is always solvable over $\mathbb{R}$. We now turn to the complex case. By unitary triangulation, we may assume that $C$ is upper triangular. As $C$ is traceless, if it has some nonzero diagonal entries, some two of them must be distinct. Again, assume that they are $c_{11}$ and $c_{22}$. Perform diagonalization on the leading 2-by-2 principal block of $C$, we may further reduce it to a diagonal 2-by-2 block. Now we have
$$
\frac{1}{1+z^2}\begin{pmatrix}1&-z\\ z&1\end{pmatrix}
\begin{pmatrix}c_{11}&0\\0&c_{22}\end{pmatrix}
\begin{pmatrix}1&z\\-z&1\end{pmatrix}
=\frac{1}{1+z^2}\begin{pmatrix}c_{11}+c_{22}z^2&\ \ast\\ \ast&\ \ast\end{pmatrix}.
$$
As $c_{11}$ and $c_{22}$ are nonzero and distinct, $c_{11}+c_{22}z^2=0$ is always solvable for some $z\in\mathbb{C}$ with $z^2\not=-1$. Therefore, in both the real and complex cases, the $(1,1)$-th entry of $C$ can be made zero via a certain similarity transform. Continue in this manner recursively for the trailing principal submatrices of $C$, we obtain a matrix with zero diagonal.
Afternote: The above proof has made use of many properties of matrices over $\mathbb{R}$ and $\mathbb{C}$, so I am not sure whether the idea of proof is applicable when the ground field is different. If not, clearly some people will find the proof dissatisfactory because it does not reveal the true reason why the set of commutators is a matrix subspace. This is reminiscent of the Cayley-Hamilton Theorem for real matrices, for which we can embed the ground field into $\mathbb{C}$ and prove the theorem easily for those unitarily diagonalizable matrices first, then use a continuity argument to finish the proof. Algebraists usually regard this proof as dissatisfactory, but those who mostly work on $\mathbb{R}$ and $\mathbb{C}$ may take a different view. At any rate, the following reference contains a relatively short proof (which fills two and a half pages) of the statement that every traceless matrix over a general field is a commutator:
A.A. Albert and Benjamin Muckenhoupt (1957), "On matrices of trace zero". Michigan Math. J., 4(1):1-3.
For (c), do you know about generalized eigenspaces?
In general, it is hard to give an answer without knowing what you know. There is a theorem saying $A$ is similar to an upper triangular matrix if and only if the characteristic polynomial of $A$ splits into linear factors; if you know about that, then you can answer (b) easily. And even if you don't know that theorem, there may be some other thing you do know from which it can be deduced.
Best Answer
For (1), see the citation in my answer to a previous question. In particular, yes, the set of all traceless matrices are precisely the set of all commutators, regardless of the underlying field.
The exercise in Hoffman and Kunze asks whether the subspace of all traceless matrices is equal to the subspace spanned by all commutators. This is different from asking whether the subspace of all traceless matrices is equal to the set of all commutators. Put it another way, the exercise in Hoffman and Kunze evades the question of whether all commutators form a matrix subspace.
For (2), see my aforementioned answer again.
For (3) and (4), consider $I_2$ over $\mathbb{F}_2$.