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
Let $F$ be any set satisfying conditions 1-3. Enumerate all open balls (or your favourite countable basis of open sets) in the interval as $(B_n)_n$, and put $\mathcal U:=\{B_n\mid B_n\cap F\textrm { is countable}\}$, and let $F':=F\setminus \bigcup \mathcal U$ (this is called the perfect (or Cantor-Bendixson) kernel of $F$). Check that $F'$ satisfies your conditions (note that $\bigcup \mathcal U\cap F$ is a countable open subset of $F$).
This is a special case of a general theorem called the Cantor-Bendixson theorem: any closed subset of a Polish (i.e. separable completely metrisable) space can be written as a disjoint union of a perfect set (i.e. closed, without isolated points) and a countable set.