Set-theory confusing a bit 
2 subsets of $\mathbb R $ are given:
$(0,\frac{2}{e}), (0,+\infty)$
• Find $A$, where $A=B\cup C.$ Where, $B=(0,\frac{2}{e}),C= (0,+\infty)=\mathbb R^*_+$.

Personal work:
Well, $$(0,\frac{2}{e})\in\mathbb R^*_+$$ but that means that $B$ is basically $C-\frac{2}{e}$ so, $A=B=\mathbb R^*_+-\frac{2}{e}\Rightarrow A=B \cup C=\mathbb R^*_+-\frac{2}{e}$. Is this correct?
 A: Just do it:
$B = (0, \frac 2e) = \{x\in \mathbb R| 0 < x < \frac 2e\}$.
$C = (0, +\infty) = \{x \in \mathbb R| 0 < x\}$.
So $A = B \cup C = \{x \in \mathbb R| 0 < x < \frac 2e$ or $0 < x\}$.
The statement $0 < x < \frac 2e$ is conjunction of two statements i) $0 < x$ and ii) $x < \frac 2e$ so we want:  $(0 < x$ and $x < \frac 2e)$ or $(0< x)$
So you have a conjunction of two statements; (i and ii) or (i)$
If i) is false then "(i and ii)" is false and "(i and ii) or (i)" is false.  So i) may not be false (whether or not ii) is true of false). 
If i) is true then "(i)" is true and "(i and ii) of (i)" is true whether or not (ii) is false.
So the upshot is "(i and ii) or (i)" is true iff and only if i) is true.
So 
$A = B \cup C = \{x \in \mathbb R| 0 < x < \frac 2e$ or $0 < x\}=\{x \in \mathbb R| 0< x\} = (0, +\infty)$
... that's it.
Now, to critique your work:
"but that means that $B$ is basically $C−\frac 2e$."
Well, no....  $1 > \frac 2e$ so $1 \not \in B$ but $1 \in C$ so $\frac 2e$ is not the only thing that needs to be removed from $C$. $1$ must be removed from $C$ as well.  And $2,3,4,5,6.... > \frac 2e$ must also be removed from $C$.  And for any $y \ge \frac 2e$ then $y$ must be removed.  SO the ENTIRE $[\frac 2e,+\infty)$ must be removed from $C$.
But if you are comfortable with the notation: you can say $B = C$ setminus $[\frac 2e, +\infty)$.
And $C$ setminus $[\frac 2e, +\infty)= $
$\{x\in C|$ it is not that case that $x \ge \frac 2e\} = $
$\{x\in C| x < \frac 2e\} = $
$C \cap \{x\in \mathbb R|x < \frac 2e\} = $
$\{x\in \mathbb R|x > 0\} \cap  \{x\in \mathbb R|x < \frac 2e\}=$
$\{x\in \mathbb R|x > 0$ and $x < \frac 2e\}=$
$\{x \in \mathbb R|0 < x < \frac 2e\} = (0, \frac 2c\}$.
"so, A=B=R∗+−2e⇒A=B∪C=R∗+−2e."
What.... I can't.... follow at all.  You believe $B = C - $ a point so $A = B \cup C = $ ..... $B$????  How on earth does that follow? and form that you get that $B$ equals all of $\mathbb R$ minus a point????  I.... don't get it.
====  OR THE MIND BOGGLING OBVIOUS WAY TO DO IT =====
If $X \subset Y$ then $Z = X\cup Y = Y$.
So $B= (0, \frac 2e) \subset (0, +\infty) = C$.
So $A = B \cup C = C$.
