1.34

I'll prove such a function exists by actually constructing one.

Let $F:[0,1] \to [0,1]$ such that:

$F(x) = \Bigg\lbrace \begin{array}{cc} f(x) & x \in \mathcal{C}_\epsilon & \\ g(x) & x \in [0,1]\cap \mathcal{C}_\epsilon^c \\ \end{array}$

Our mapping $f(x):\mathcal{C}_\epsilon \to \mathcal{C}_\gamma$ is constructed by simply reassinging $x \in \mathcal{C}_\epsilon$'s binary expansion to $\mathcal{C}_\gamma$'s. To do this, simply observe that for any $x$ in $\mathcal{C}_\epsilon$,

$x = \sum_{n=1}^{\infty} \alpha_n \epsilon^n$ where $\hspace{0.25cm} \alpha_n \in \lbrace 0, 2 \rbrace$, and is unique to $x$.

Thus,

$f(x) = \sum_{n=1}^{\infty} \alpha_n \gamma^n$

Next, we need to construct $g(x)$. These functions are going to be continuous (linear) functions that map open disjoint intervals to corresponding open disjoint intervals.

First, lets define the set $\beta_{n,k}$ as the "binary representation of the integer $k$ with a string of length $n$". For example: $$\beta_{5,3} = (0,0,1,0,1)$$ Next, augment $\beta$ with a $1$ on the rightmost-side of the string. Call this set $\phi_{n,k}$. I.e. $$\phi_{5,3} = (0,0,1,0,1,1)$$ Now, let $\phi_{n,k}(i)$ be the $i^{th}$ element of the string, from the left.
(I.e. $\phi_{5,3}(3) = 1$.)

Moving forward, we'll use $\phi_{n,k}$ to define the sets $\mathcal{O}_{n,k}^\epsilon$ (and similarly for $\mathcal{O}_{n,k}^\gamma$) such that: $$\bigcup_{n = 0}^\infty \bigcup_{k = 0}^{2^n - 1} \mathcal{O}_{n,k}^\epsilon = \hspace{0.25cm} [0,1]\cap \mathcal{C}_\epsilon^c$$ Without further ado, $$\mathcal{O}_{n,k}^\epsilon = \Bigg( \sum_{i=1}^{n+1} \Big(\frac{(1-\epsilon)^{i-1}(1+\epsilon)}{2^i}\phi_{n,k}(i)\Big)-\epsilon \Big(\frac{(1-\epsilon)}{2}\Big)^n, \sum_{i=1}^{n+1} \Big(\frac{(1-\epsilon)^{i-1}(1+\epsilon)}{2^i}\phi_{n,k}(i)\Big) \Bigg)$$ While the expressions are regrettably unpleasant, notice that for each $n,k, \epsilon$, it successfully reproduces the correct $(k+1)^{th}$ open interval for the $(n+1)^{th}$ generation of an $\epsilon$-cantor set. For example, we know that the second interval in the third generation of the standard ($\epsilon = \frac{1}{3}$) Cantor set is $\big(\frac{7}{27}, \frac{8}{27}\big)$. We can easily check...

First, lets calculate: $$\sum_{i=1}^{4} \Big(\frac{(\frac{2}{3})^{i-1}(\frac{4}{3})}{2^i}\phi_{3,2}(i)\Big) = \frac{2}{9} + \frac{2}{27} = \frac{8}{27}$$ With that out of the way, it's easy to check now that: $$\mathcal{O}_{2,1}^{\frac{1}{3}} = \Big(\frac{7}{27}, \frac{8}{27} \Big)$$ So... it works! This was constructed so that the collection of all $\mathcal{O}_{n,k}^\epsilon$'s are pairwise-disjoint. From here, it's actually easy to define a linear function from $\mathcal{O}_{n,k}^\epsilon$ $\to$ $\mathcal{O}_{n,k}^\gamma$. The slope from one set to the other is (obviously) $\big(\frac{\gamma}{\epsilon}\big)^{n+1}$. The $y$-intercept $b_{n,k}$ is fairly straightforward to calculate... (just pick the middle points of the corresponding intervals, and evaluate.)

Thus, we can finally fully define $g$ as follows: $$g(x) = \sum_{n=0}^{\infty} \sum_{k = 0}^{2^n - 1} \Big(\big(\frac{\gamma}{\epsilon}\big)^{n+1} x + b_{n,k}\Big)\chi_{\mathcal{O}_{n,k}^\epsilon}$$ Finally, we have our function $F(x) = f(x) + g(x)$, which satisfies all three properties outlined in the exercise.