EDIT: It looks like I was too hasty to delete this answer. I think I have fixed the gap that Anton noted in the comments and in his answer. As Anton pointed out in his answer, the trace loop that I dropped out at the end is nontrivial, but it can be removed from any diagram where an "X" appears in another connected component. Intuitively, an idempotent should have dimension 1 wherever it is nonzero. I've oriented the string diagrams vertically so that any horizontal cross-section can be read left-to-right in symbolic order.

The answer to question 2 (hence 3) is "yes" for symmetric monoidal categories. Let $f:X\otimes X\to X$ be the two-sided inverse of $id_x\otimes i$. As it turns out, $f$ is also the left inverse of $i\otimes id_X$. This is proven in the image linked below.


Let $\phi:X\to X^\vee$ be the composition 

$X\to X\otimes I\to X\otimes X\otimes X^\vee\to X\otimes X^\vee\to I\otimes X\otimes X^\vee\to X^\vee\otimes X\otimes X\otimes X^\vee\to X^\vee\otimes X\otimes X^\vee\to X^\vee\otimes I\to X^\vee,$

and let $\psi:X^\vee\to X$ be the composition $X^\vee\to I\otimes X^\vee\to X\otimes X^\vee\to I\to X$. These two maps are mutually inverse.
[Here is a string diagram "proof" of both of the above facts.][1]


[This is the old proof with an unjustified step at the end.][2]


  [1]: https://i.sstatic.net/FDNYK.jpg
  [2]: https://i.sstatic.net/HWgBQ.jpg