The answer to question Q1 is no. The answer to question Q2 is yes, though it becomes no if we replace "differentiable" with "continuously differentiable".
The key reference is the following:
Dovgoshey, O., et al. “The Cantor function.” Expo. Math. 24 (2006). 1–37
The following theorem appears on pg. 25 of the paper (Proposition 7.5):
Theorem 1. Let $g\colon[0,1]\to[0,1]$ be the Cantor function. Then there exists a homeomorphism $\varphi\colon[0,1]\to[0,1]$ so that $g\circ\varphi$ is everywhere differentiable, with uniformly bounded derivative.
Immediately afterwards (Proposition 7.6), it is proven that:
Theorem 2. Let $g\colon[0,1]\to[0,1]$ be the Cantor function. Then there do not exist homeomorphisms $\varphi,\psi\colon[0,1]\to[0,1]$ so that $\psi\circ g\circ\varphi$ is continuously differentiable.
This settles question Q2. Also, it follows from Theorem 1 that the answer to question Q1 is no. In particular, if $C$ is the Cantor set, then $C$ is nowhere dense, so $\varphi^{-1}(C)$ is nowhere dense as well. But the image of $\varphi^{-1}(C)$ under the differentiable function $g\circ\varphi$ is the same as $g(C)$, which is the entire unit interval $[0,1]$.
Of course, this does not address the question of whether the image of a nowhere dense set under a continuously differentiable function is always nowhere dense. The post linked to by Henno is not entirely convincing, partially because I cannot follow the proof, but mostly because the author seems to retract the proof later in the thread:
Though it's not entirely clear, the conclusion at the end of the thread seems to be that the answer to Q1 is no even in the $C^1$ case. (Edit: As Landscape and George Lowther point out, there is indeed a simple argument that the answer to Q1 is yes for $C^1$ functions.)
Note that this answer does not conflict with this post, which states that the image of a set of measure zero under a differentiable function has measure zero. Assuming the post is correct, the inverse image $\varphi^{-1}(C)$ of the Cantor set under the homeomorphism $\varphi$ must have positive Lebesgue measure. This is of course possible for a nowhere dense set, e.g. the fat Cantor set.
Best Answer
Regarding your confusion for the converse part, $A$ is dense in the non-empty open set $G = (0,1)$. This is because by the definition of denseness, $A$ is dense in $G$ iff $A \cap G$ is dense in $G$. That is, iff $G \cap (\overline{A \cap G}) = G$. Here, $A \cap G = G$, and $\overline{A \cap G} = A$, so $G \cap (\overline{A \cap G}) = G$.
Additionally, the proof you have given for the forward implication is slightly incorrect. You say that $(\bar{A})^0 = \emptyset \implies \bar{A}$ is not a nonempty open set. But this is assuming too much; the correct implication is: $(\bar{A})^0 = \emptyset \implies \bar{A}$ does not contain any nonempty open set.
To correct the proof, one could work as follows. Let $(\bar{A})^0 = \emptyset$. Then, $\bar{A}$ does not contain any nonempty open subset of $X$. Let $G$ be a nonempty open subset of $X$. If $A$ is dense in $G$, then $$ G \cap (\overline{A \cap G}) = G. $$ But, $$ A \cap G \subset A \implies \overline{A \cap G} \subset \bar{A}. $$ Therefore, $G \cap (\overline{A \cap G}) \subset G \cap \bar{A} \subset G$. Hence, $$ G \cap (\overline{A \cap G}) = G \implies G \cap \bar{A} = G \implies G \subset \bar{A}. $$ But this contradicts that $\bar{A}$ does not contain any nonempty open subset of $X$. Hence, $A$ is not dense in $G$.
Further comments: Your proof breaks down at the point where you conclude from $\bar{A}$ not being a (nonempty) open set to the fact that $A$ is not dense in any nonempty open subset of $X$. In general this is not true.
To see why, we can take the same sets you considered for the reverse implication. Let $X = \mathbb{R}$, $A = [0,1]$. Then $\bar{A} = A$, which is not a nonempty open set. But $A$ is still dense in the set $G = (0,1)$. So, just having that $\bar{A}$ is not a nonempty open set does not imply that $A$ is not dense in any nonempty subset of $X$.