A Cantor set $C\subset[0,1]$ is closed, and that is all we need.
Therefore $f(x)=d(x,C)$ (distance from $x$ to $C$) vanishes on and only on $C$.
It is also Lipschitz-continuous.

For your third question, you can take $g(x)=f(x)^2$.
This is $C^1$ apart from peaks in the middle points of the intervals of the complement of $C$.
You can easily smooth the function near these points, since you are well away from $C$.

Choosing a higher power gives you any $C^k$ smoothness you want.
But you can also get $C^\infty$.
Let $\phi:\mathbb R\to[0,\infty)$ be a smooth function vanishing on $(-\infty,0]$ and set $g(x)=\phi(f(x))$ and mollify the tips.

Conclusion: The answer to all three questions is yes.

**Edit:**
It is not entirely clear what you mean by constructing by hand.
This is as close as I can come to making an explicit function if the Cantor set is not specified.
(I also noted that the distance was also used in this answer to the OP's linked question. But it was not concerned with regularity.)