Let $B$ be a Borel measurable set, and let $f$ be a continuous strictly increasing function, then $f^{-1}(B)$ is a Borel set.
As t.b. said, for this part you don't need $f$ to be strictly increasing: the statement holds for all continuous functions. See, for example,
https://math.stackexchange.com/q/891568/
Strictly increasing continuous functions map Borel sets to Borel sets
To handle this question (stated in the title), consider the inverse function $g=f^{-1}$. It is a continuous function, defined on the range of $f$ (an interval of $\mathbb R$). Apply the first part to $g$.
You can construct $f$ as follows: let $m$ denote Lebesgue measure. Once $m(A)=0$, we can choose for each $i=1,2,\cdots$, a sequence of intervals $I_{ij}$ such that $$A\subset \cup_{j=1}^\infty I_{ij},\ \sum_{j=1}^\infty m(I_{ij})\le \frac{1}{2^i}. \tag{1}$$
Note that each $x$ belongs to infinitely many intervals $I_{ij}$. Define $f:[0,1]\to\mathbb{R}$ by $$f(x)=\sum _{i,j=1}^\infty m(I_{ij}\cap [0,x]).$$
Note that $f$ is increasing. Also $f$ satisfies $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\infty,\ \forall x\in A.$$
Indeed, as $x$ belongs to infinitely many intervals $I_{ij}$, we can assume without of generality that $x$ belongs to $J_k=(x-a_k,x+a_k)$ for all $k=1,2,\cdots$, that each $J_k=I_{ij}$ for some $i,j$ and $a_k\to 0$. Note that
\begin{eqnarray}
\frac{f(x-h)-f(x)}{h} &=& \frac{1}{h}\sum _{i,j=1}^\infty m(I_{ij}\cap [x-h,x]) \nonumber \\
&\ge& \frac{1}{h}\sum _{k=1}^\infty m(J_k\cap [x-h,x]) \nonumber \\
&\ge& \frac{1}{h}(Nh+\sum _{k=N+1}^\infty a_k),
\end{eqnarray}
where $N$ depending on $h$ is choosen in such a way that $h\le a_N$. As $h\to 0^+$, we see from the previous inequality that the left derivative is infinity. With an similar reasoning, you can conclude that the right derivative is also infinity.
To prove that $f$ is absolutely continuous, note that $$\sum _{k=1}^N |f(x_{k+1})-f(x_k)|=\sum _{k=1}^n \sum_{i,j=1}^\infty m(I_{ij}\cap[x_k,x_{k+1}]). \tag{2}$$
Once $(1)$ is satisfied, you conclude from $(2)$ that $f$ is absolutely continuous. To get a strictly increasing functions, just consider $g(x)=x+f(x)$.
Note: The original construction is due to professor Porter and it is contained here.
Best Answer
Note that $$\frac{F(x)-F(x+h)}{h}=\frac{m(A\cap [x+h,x])}{h}\le 2\frac{m(A\cap[x-h,x+h])}{2h},\ \forall h<0 \tag{1},$$
and
$$\frac{F(x+h)-F(x)}{h}=\frac{m(A\cap [x,x+h])}{h}\le 2\frac{m(A\cap[x-h,x+h])}{2h},\ \forall h>0 \tag{2},$$
Lebesgue density theorem, $(1)$ and $(2)$ implies that $$ F'(x)=0,\ a.e. \ x\in [0,1]\setminus A.$$
Edit: As @Dave pointed out in the comment, we also have that $F'(x)=1$ a.e. $x\in A$. This implies in particular that $G'(x)=1$ a.e. $x\in A$ and $G'(x)=-1$ a.e. $x\in [0,1]\setminus A$, therefore, by using the properties of $A$, it follows immediate that $G$ cannot be monotone in any interval.