How to find the values of $m$ so $A \cap B = \emptyset$ and $A \cup B = (-3,+\infty)$?

Question

I have two intervals $A = (-3, m^2- 2]$ and $B=[m, +\infty)$. How to find the values of $m$ so that $A \cap B = \emptyset$ and $A \cup B = (-3,+\infty)$? I tried

{a = x > -3 && x < m^2 - 2, b = x >= m} {Reduce[a || b]}

and

{a = x > -3 && x < m^2 - 2, b = x >= m}
{Reduce[a && b]}

We have $A \cap B = \emptyset$ when and only when $$\begin{cases} m^2 - 2 > -3&\\ m^2 - 2 <m. \end{cases}$$ and $A \cup B = (-3,+\infty)$ when and only when $$\begin{cases} m^2 - 2 > -3&\\ m^2 - 2 \geqslant m. \end{cases}$$ Therefore we need to solve two systems of inequalities.

Answer 1

Just reduce the four inequalities you gave:

Reduce[m^2 - 2 > -3 && m^2 - 2 < m && m^2 - 2 > -3 && m^2 - 2 >= m, m]

(* False *)

Your second and fourth inequalities contradict each other, so there is no solution.