Is a r.v. supported on $\mathbb Q$ discret or continuous ? Same question for $\mathbb R\backslash \mathbb Q$. I know that a discret r.v. is a r.v. that is supported on a finite set or a countable set. But if $X$ is s.t. $\mathbb P\{X\in\mathbb Q\}=1$, how should I right the expected value :
$$\mathbb E[X]=\sum_{x\in\mathbb Q}x\mathbb P\{X=x\}$$
or $$\mathbb E[X]=\int_{\mathbb Q}x f_X(x)dx\ \ ?$$
The sum doesn't has sense for me, so I would say that the integral is the correct formulae. But if it is, is really $X$ a discrete r.v. ?
Same question if $\mathbb P\{X\in \mathbb R\backslash \mathbb Q\}=1$. To me s.t. a r.v. is not continuous, but is not continuous as well. So what can it be ?
There is nothing wrong with the sum. It looks strange because you don't usually think of $\mathbb{Q}$ as being well ordered, but that's just a small flaw. For the sum to make sense you will of course need it to converge absolutely and then it will converge independently of how you choose the well ordering of $\mathbb{Q}$. 

(re.: acetone's comment) The definition of the expected value for a countably infinitely supported r.v. contains the requirement that the sum converge absolutely.

As to the second question in the OP. You don't point out what the probability space here is or what $\mathbb{P}$ is and for continuous random variables that is quite important. 

But for example take $X$ a continuous random variable with a normal distribution on the reals. Then $\mathbb{P}(\{X\in\mathbb{R}\setminus\mathbb{Q}\})=1$ will be true since $\mathbb{Q}$ has Lebesgue measure $0$.

Again following from helpful comments by acetone:

Continuous random variable might be the part that's confusing you OP. Depending on what your definition of that term is you get very different results. In general you have discrete random variables (those are on probability spaces with finite or countably infinite many elements), continuous random variables those have a continuous probability distribution function (usually we think of them as having values in some subset of the reals with the corresponding borel $\sigma$-algebra as the measurable sets), and then there the "other" random variables.

Sometimes the "other" random variables get lumped up into the continuous random variables because they are not discrete and we think of the underlying sets (Reals, Complex numbers) as being continuous in some sense but that leads to confusion.
There is nothing wrong with the sum. It looks strange because you don't usually think of $\mathbb{Q}$ as being well ordered, but that's just a small flaw. For the sum to make sense you will of course need it to converge absolutely and then it will converge independently of how you choose the well ordering of $\mathbb{Q}$. 

(re.: acetone's comment) The definition of the expected value for a countably infinitely supported r.v. contains the requirement that the sum converge absolutely.

As to the second question in the OP. You don't point out what the probability space here is or what $\mathbb{P}$ is and for continuous random variables that is quite important. 

But for example take $X$ a continuous random variable with a normal distribution on the reals. Then $\mathbb{P}(\{X\in\mathbb{R}\setminus\mathbb{Q}\})=1$ will be true since $\mathbb{Q}$ has Lebesgue measure $0$.

Again following from helpful comments by acetone:

Continuous random variable might be the part that's confusing you OP. Depending on what your definition of that term is you get very different results. In general you have discrete random variables (those are on probability spaces with finite or countably infinite many elements), continuous random variables those have a continuous probability distribution function (usually we think of them as having values in some subset of the reals with the corresponding borel $\sigma$-algebra as the measurable sets), and then there the "other" random variables.

Sometimes the "other" random variables get lumped up into the continuous random variables because they are not discrete and we think of the underlying sets (Reals, Complex numbers) as being continuous in some sense but that leads to confusion.