The cleanest way to do this that I know is to show that $F$ is the uniform limit of a sequence of continuous functions, and is hence continuous.
Consider the sequence $C_k$ of closed sets which go in to the construction of the Cantor set $C$ (ie. $C_k$ is the union of closed intervals that are remaining at the $k^{th}$ stage). Thus
$$
C = \cap_{k=1}^{\infty} C_k
$$
Now let $D_k = [0,1]\setminus C_k$ which is a union of $2^k - 1$ intervals $\{I_j^k\}$, ordered from left to right. Define $f_k:[0,1]\to \mathbb{R}$ be the continuous function defined such that
$$
f_k(0) = 0, \qquad f_k(1) = 1
$$
$$
f_k(x) = \frac{j}{2^k} \quad \forall x\in I_j^k, 1\leq j \leq 2^k -1
$$
and which is linear on each interval of $C_k$. (Try drawing the pictures for $f_1$ and $f_2$ to get an idea as to what is going on).
Now $\{f_k\}$ is monotone increasing, and
$$
f_{k+1} = f_k \quad \text{on} \quad I_j^k, 1\leq j\leq 2^k -1
$$
Hence,
$$
|f_{k+1} - f_k| < 2^{-k}
$$
Hence, $\sum (f_{k+1} - f_k)$ converges uniformly on $[0,1]$, and so
$$
F = \lim_{k\to \infty} f_k
$$
is also continuous.
Edit: I just found the following picture online, which explains what $f_1$ and $f_2$ look like (The picture has an added $f_0$ function as well):
I hope the following procedure is "systematic" enough.
Start with $x\in [0,1]$. Define $a_1$ to be the greatest number in $\{ 0,1,2\}$ such that $\frac{a_1}3\leq x$; so $a_1=0$ if $x\in [0,1/3)$, $a_1=1$ if $x\in [1/3,2/3)$ and $a_1=2$ if $x\in [2/3,1]$. In each case you have
$$\frac{a_1}3\leq x\leq \frac{a_1}{3}+\frac13\,\cdot$$
Assuming $a_1,\dots ,a_{n-1}$ have already been found (for some $n\geq 2$) with $$\sum_{k=1}^{n-1}\frac{a_k}{3^k}\leq x\leq \sum_{k=1}^{n-1}\frac{a_k}{3^k}+\frac{1}{3^{n-1}}\, ,$$ define $a_n$ to be the greatest number in $\{ 0,1,2\}$ such that $\sum_{k=1}^{n-1}\frac{a_k}{3^k}+\frac{a_n}{3^n}\leq x$. Then
$$\sum_{k=1}^n\frac{a_k}{3^k}\leq x\leq \sum_{k=1}^{n}\frac{a_k}{3^k}+\frac{1}{3^{n}}\cdot $$
Indeed, if $a_n=0$ or $1$ then $a_n+1$ is still in $\{ 0,1,2\}$, so by the definition of $a_n$ you have in fact $x<\sum_{k=1}^{n-1} \frac{a_k}{3^k}+\frac{a_n+1}{3^n}=\sum_{k=1}^n\frac{a_k}{3^k}+\frac{1}{3^n}$; and if $a_n=2$ then you may write (by the induction hypothesis) $x\leq \sum_{k=1}^{n-1}\frac{a_k}{3^k}+\frac{1}{3^{n-1}}=\sum_{k=1}^{n-1} \frac{a_k}{3^k}+\frac{2}{3^{n}}+\frac{1}{3^n}=\sum_{k=1}^n\frac{a_k}{3^k}+\frac1{3^n}\cdot$
So, constructing your sequence in this way, you obviously get $x=\sum_{k=1}^ \infty \frac{a_k}{3^k}\cdot$
As you know, the ternary expansion is not necessarily unique. If you start with a number $x$ of the form $x=\sum_{k=1}^N\frac{b_k}{3^k}$ (finite sum), then the procedure gives you $a_1=b_1, \dots , a_N=b_N$ and $a_k=0$ for all $k>N$.
As for your second question, you already have a nice answer.
Best Answer
Remember that $a_n-b_n\in\{-2,0,2\}$.
Suppose that $$ \sum_{n=1}^\infty\frac{a_n-b_n}{3^n}\gtrless0\tag{1} $$ $(1)$ happens if and only if, for the smallest $n_0$ so that $a_{n_0}\not=b_{n_0}$, it is the case that $a_{n_0}\gtrless b_{n_0}$.
The claim above is true because $$ \text{If }a_{n_0}>b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{3^n}\ge\frac2{3^{n_0}}-\sum_{n>n_0}\frac2{3^n}=\frac1{3^{n_0}}\tag{2} $$ $$ \text{If }a_{n_0}<b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{3^n}\le-\frac2{3^{n_0}}+\sum_{n>n_0}\frac2{3^n}=-\frac1{3^{n_0}}\tag{3} $$ If for the smallest $n_0$ so that $a_{n_0}\not=b_{n_0}$, it is the case that $a_{n_0}\gtrless b_{n_0}$, then $$ \sum_{n=1}^\infty\frac{a_n-b_n}{2^{n+1}}\gtreqless0\tag{4} $$ This is because $$ \text{If }a_{n_0}>b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{2^{n+1}}\ge\frac1{2^{n_0}}-\sum_{n>n_0}\frac1{2^n}=0\tag{5} $$ $$ \text{If }a_{n_0}<b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{2^{n+1}}\le-\frac1{2^{n_0}}+\sum_{n>n_0}\frac1{2^n}=0\tag{6} $$ So, if $x\gtrless y$, $F(x)\gtreqless F(y)$. We can't get strict increase since $F$ is constant on all the "middle-thirds" intervals.