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.
First, the answer to question 1 is "no". Take $C$ to be the category of $k$-vector spaces with the monoidal structure induced by the tensor product, and let $X$ be the countably infinite direct sum of $k$ with itself. Then $X\cong X\otimes X$, but its dual has uncountable dimension.
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.
