For (b), the key observation is that, as $g$ is continuous, the image of an interval is an interval, and so $m(g((a,b)))=g(b)-g(a)$.
Then, as Tim said, we can use that the complement of $C$ is a countable disjoint union of open intervals $C_k$ whose measures add to 1, and that on these intervals we have $f=0$. So, if $C_k=(a_k,b_k)$, then
$$
m(g(C_k))=g(b_k)-g(a_k)=f(b_k)-f(a_k)+b_k-a_k=b_k-a_k=m(C_k).
$$
So
\begin{align}
2-m(g(C))&=m([0,2]\setminus g(C))=m(g(\bigcup_k C_k))=m(\bigcup_k g(C_k))\\ \ \\ &=\sum_k m(g(C_k))
=\sum_k m(C_k)=1,
\end{align}
i.e. $m(g(C))=1$.
For (c), it is an established result that any set of positive measure contains a non-measurable subset. So let $A\subset g(C)$ be non-measurable, and put $B=g^{-1}(A)$. Since $B\subset C$ and $C$ is a null-set, $B$ is also a null-set, and it particular it is measurable. But it is not Borel: since $g^{-1}$ is continuous, the pre-image of a Borel set is Borel; as $A=g(B)$, this would make $A$ Borel and thus measurable, which it's not.
Finally, for (d), the key is that
$$
1_A=1_{g(B)}=1_B\circ g^{-1}.
$$
The characteristic $1_A$ is not measurable since $A$ isn't. But $B$ is measurable so $1_B$ is, and $g^{-1}$ is continuous. So the composition of measurable functions may fail to be measurable, even if one of them is continuous.
A function $f:\mathbb R\to\mathbb R$ is Borel if and only if for all $a\in\mathbb R$, the set $\{x\in\mathbb R:f(x)>a\}$ is a Borel subset of $\mathbb R$.
Suggestion: Think about what possible types of sets you can get for $\{x\in\mathbb R:f(x)>a\}$ when $f$ is a monotone function. You may want to conjecture with the aid of examples before trying to prove your conjecture. The sets you get should be easily confirmed to be Borel.
Best Answer
Here's an attempt to salvage Matthew Pancia's solution, which unfortunately depended on an uncountable union over all possible derivatives.
Given $f$ we can define, in the obvious way, a continuous function $F:\mathbb R\times(\mathbb R\setminus 0)\to \mathbb R$ such that $f$ is differentiable at $x$ exactly when $\lim_{h\to 0} F(x,h)$ exists. The usual formalization of this is $$\exists y:\forall\varepsilon:\exists \delta:\forall h: |h|<\delta\Rightarrow |F(x,h)-y|<\varepsilon$$ Classically all of the variables here are real, but it is easy to see that we can restrict $\varepsilon$ and $\delta$ to $\mathbb Q$ without changing the meaning. We can also restrict $h$ to $\mathbb Q$ because $F$ is continuous. However, it is essential that $y$ can be an arbitrary real, because otherwise we would be looking for points where $f$ is differentiable with rational derivative, which is something quite different.
However, we can also formalize the existence of a limit like $$\forall\varepsilon:\exists \delta:\exists Y:\forall h: |h|<\delta \Rightarrow |F(x,h)-Y|<\varepsilon$$ This works because $\mathbb R$ is complete; it is essentially the same as changing "has a limit" about a sequence to "is Cauchy". The arguments that $\varepsilon$, $\delta$ and $h$ can be restricted to the rationals work as before, but now $Y$ can also be taken to be a rational in each case.
For each particular choice of $\varepsilon$, $\delta$, $Y$, and $h$, the set of $x$ such that $|h|<\delta\Rightarrow |F(x,h)-Y|<\varepsilon$ is open and therefore Borel.
Now handle each of the quantifiers from the inside out: For each choice of $\varepsilon$, $\delta$, and $Y$, the set of $x$ such that $$\forall h: |h|<\delta \Rightarrow |F(x,h)-Y|<\varepsilon$$ is a countable intersection of Borel sets and therefore Borel. For each choice of $\varepsilon$ and $\delta$ the set of $x$ such that $$\exists Y:\forall h: |h|<\delta \Rightarrow |F(x,h)-Y|<\varepsilon$$ is a countable union of Borel sets and therefore Borel. And so forth. At the top we find that the set of points of differentiability is Borel and thus in particular measurable.