Solve the inequality $x + 2 < -8$ for the following cases: $(1) x \in N$,$(2) x \in Z$,$(3) x \in R$.

(N/A) Given the inequality: $x + 2 < -8$.
Subtracting $2$ from both sides: $x < -8 - 2$,which simplifies to $x < -10$.
$(1)$ For $x \in N$ (natural numbers): Since there are no natural numbers less than $-10$,the solution set is $\phi$ (the empty set).
$(2)$ For $x \in Z$ (integers): The integers less than $-10$ are $\{\ldots, -13, -12, -11\}$.
$(3)$ For $x \in R$ (real numbers): The solution set is the interval $(-\infty, -10)$.