TL:
$a) \sqrt{2x} + 3 = 1 +$ $\sqrt{2}$
$⇒ \left \{ {{2x + 3 = ( 1 +\sqrt{2})^2} \atop {2x+3>0}} \right.$
$⇒ \left \{ {{2x+3 = 3+2\sqrt{2}} \atop {x>\frac{2}{3}}} \right. => x =\sqrt{2} $
$b)\sqrt{10 + \sqrt{3x}} = 2\sqrt{6} $
$⇒ 10 +\sqrt{3x} = 10+4\sqrt{6}$
$⇒ \sqrt{3x} = 4\sqrt{6}$
$⇒ 3x=96 $
$⇒ x= 32$