Given two any intervals $$J$$ and $$K$$ we will say that $$J$$ is contained in $$K$$, $$J\subseteq K$$, if all the elements of $$J$$ belong to $$K$$.
The interval $$[3,7]$$ is included in the interval $$\Big(-2,\dfrac{15}{2}\Big)$$, and we denote it by: $$$[3,7] \subseteq \Big(-2,\dfrac{15}{2}\Big)$$$ since $$3 \in \Big(-2,\dfrac{15}{2}\Big)$$, $$7 \in \Big(-2,\dfrac{15}{2}\Big)$$ and, consequently, for any $$x \in [3,7]$$ it is satisfied that $$x \in \Big(-2,\dfrac{15}{2}\Big)$$
Intuitively, we'd say that this is an order because, given two intervals, it shows which one is bigger: if $$J\subseteq K$$ then $$J$$ is smaller than $$K$$.
In contrast to the order on the real numbers it is not a total order, that is, not every pair of intervals are comparable.
Considering the intervals $$(2,3)$$ and $$(3,4)$$, let's see that they are not comparable.
$$\dfrac{5}{2}\in (2,3),$$ but $$\dfrac{5}{2}\notin (3,4),$$ therefore it is not true that $$(2,3)\subseteq (3,4).$$
Likewise,
$$\dfrac{10}{3}\in (3,4),$$ but $$\dfrac{10}{3}\notin (2,3),$$ therefore it is not true either that $$(3,4)\subseteq (2,3).$$
So, they are not comparable.