You should get that the answer is $1/2$, immediately from integration by parts. See this, in particular the sections "Integration by parts" and "Related concepts". Further, see this for a proof of the integration by parts formula.
EDIT: Explicitly, since $f$ is continuous and non-decreasing, and is constant on $\mathbb{R}-[0,1]$, $\int_{\mathbb R} {f(x)du(x)} = \int_0^1 {f(x)df(x)} $, and it holds
$$
\int_0^1 {f(x)df(x)} = f(1)f(1) - f(0)f(0) - \int_0^1 {f(x)df(x)}
$$
(integration by parts). Since $f(0)=0$ and $f(1)=1$, it thus follows that $\int_{\mathbb R} {f(x)du(x)} = 1/2$.
EDIT: More generally, if $F$ is any continuous distribution function (the Cantor function $f$ is a particular example), then $\int_{\mathbb R} {F(x)dF(x)} = 1/2$. As before, this can be proved using integration by parts, which is allowed since $F$ is continuous and non-decreasing. (Under certain conditions, this also follows from a change of variable.) Indeed, for any $a < b$,
$$
\int_a^b {F(x)dF(x)} = F(b)F(b) - F(a)F(a) - \int_a^b {F(x)dF(x)}.
$$
Hence the result follows by letting $a \to -\infty$ and $b \to \infty$.
Another way to obtain the general result is as follows. Let $X$ be an arbitrary random variable with continuous distribution function $F$. Then, $\int_{\mathbb R} {F(x)dF(x)}$ expresses the expectation of the random variable $F(X)$. As is well known, in this case $F(X) \sim {\rm uniform}(0,1)$. Hence, $\int_{\mathbb R} {F(x)dF(x)} = 1/2$.
Remark. With $f$ as above, integration by parts gives
$$
\int_0^1 {xdf(x)} = xf(x)\big|_0^1 - \int_0^1 {f(x)dx} = 1 - 1/2 = 1/2.
$$
The left-hand side, $\int_0^1 {xdf(x)}$, expresses the expectation of the Cantor distribution.
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.
Best Answer
The value of the integral is $\frac{1}{2}$, by symmetry.
Notice that $\varphi(1-x) = 1 - \varphi(x)$. From this, it is easy to show that the Cantor measure $m_\varphi$ is invariant under the transformation $x \mapsto 1-x$. Thus $$\int x\, m_\varphi(dx) = \int (1-x)\, m_\varphi(dx) = 1 - \int x\,m_\varphi(dx).$$