Regular Tetrahedrons.

$a^2 + 1 = x^2 = 2(1 - a)^2 \implies a^2 + 1 = 2 - 4a + 2a^2 \implies a^2 - 4a + 1 = 0 \implies$

$a = 2 \pm \sqrt{3}$

$a = 2 + \sqrt{3} \implies 1 - a = -1 - \sqrt{3} < 0 \therefore$ drop $a = 2 + \sqrt{3}$ .

$a = 2 - \sqrt{3} \implies x = 2\sqrt{2 - \sqrt{3}} \implies A_{\triangle{AEF}} = \sqrt{3}(2 - \sqrt{3}) = 2\sqrt{3} - 3$ .

$h = \sqrt{x^2 - \dfrac{x^2}{3}} = \sqrt{\dfrac{2}{3}}x = 2\sqrt{\dfrac{2}{3}}\sqrt{2 - \sqrt{3}} \implies$

The volume of the regular tetrahedron $V_{T} = \dfrac{1}{3}(2 - \sqrt{3})\sqrt{3}(2\sqrt{\dfrac{2}{3}}\sqrt{2 - \sqrt{3}}) =$

$\dfrac{2\sqrt{2}}{3}(2 - \sqrt{3})^{\frac{3}{2}} = \dfrac{a\sqrt{a}}{b}(a - \sqrt{b})^{\frac{b}{a}} \implies a + b = \boxed{5}$ .