Does definition of an event depend on foreknowledge of probability? Assume that a dice is tossed, the event $E_1$ that the outcome is odd number can be denoted as $E_1=\{O_1,O_3,O_5\}$. We know that the event $E_2$ that the outcome is number 7 is $E_2=\{\}$ (empty set) because $O_7$ can not happen. so the definition of event $E_2$ relies on the foreknowledge that $O_7$ can not happen. otherwise we would have denoted that as $E_2=\{O_7\}$    -- it has an element albeit its probability is 0.
Should definition of an event depends on foreknowledge of probability? should not only the opposite be true? 
how to understand "empty set has no element" in event definition context?
 A: One way to overcome your problem is to define probability measure $\mathbb{P}$ over $\sigma$-Algebra of events $\mathcal{F}$ defined as subsets of some set $\Omega$.
This is one way to define events prior to defining probability used in modern probability science and I will call it a formal tool of probability space.
Probability measure is any set function satisfying $\mathbb{P}(\emptyset) = 0 $ (Probability that nothing happens is $0$),$\mathbb{P}(\Omega) = 1$ (Probability that something happens at all is $1$) and that for any disjoint countable family of events $A$ we have $\mathbb{P}(\bigsqcup_n A_n) = \sum_n \mathbb{P}(A_n)$. For such measure you can have $\mathbb{P}(A) =0$ even if $A$ is not empty. 
Depending on the peoblem you can select suitable probability space $(\Omega,\mathcal{F},\mathbb{P})$. If you want formulate questions about dice in your example involving any natural numbers we can select $\Omega = \mathbb{N}$, $\mathcal{F} = 2^\mathbb{N}$ i. e. set of all subsets of $\mathbb{N}$, and define $\mathbb{P}(\{1\}) = \frac{1}{6} \ldots \mathbb{P}(\{6\}) = \frac{1}{6}$ and for $n > 6$ $ \mathbb{P}(\{n\}) = 0 $. Using this definition you can extend $\mathbb{P}$ to all events in $\mathcal{F}$ which may involve complicated queries which actually might have zero probability. However, in this construction questions like $\mathbb{P}((1/2,3/5])$ are still not formally valid by construction of probability space.
Update:
Of course it wasn't implied in the answer that $\mathbb{P}(A) = 0$ implies that event $A$ is impossible.
Example: assume you are tossing darts into continuous interval $[0,1]$ without any aim. Eventually you hit some value, say $\sqrt{2}/3$. $\mathbb{P}(\{\sqrt{2}/3\}) = 0$, however it just has happened.
Concept of possibility seems to be mach more philosophical than mathematical. We can demand that for any true sample space $(\Omega,\mathcal{F},\mathbb{P})$ for any $\omega \in \Omega$ event $\{\omega\}$ is possible. Call this possibility property. Then think about the dice as a map (random variable)  $X : \Omega \to \mathbb{N} $. With this definition we can say that $\mathbb{P}'(X \: \mathrm{is} \: 7) = \mathbb{P}\big(X^{-1}\{7\}\big) = \mathbb{P}(\emptyset)$, hence it is impossible for the dice to take value $7$. Here we think about $\mathbb{P}'$ as a push forward of $\mathbb{P}$ to $\mathbb{N}$ through $X$.  Note, that  $(\mathbb{N},2^\mathbb{N},\mathbb{P}')$ is a probability space in Kolmogorov axiomatics but it doesn't have the possibility property.
This can be a reason that most applications of probability theory works with random variables $X:\Omega \to S $ with sample space $\Omega$ left without any explicit description.
