(a) \(2x^{2} + x - 6 < 0 \implies 2x^{2} + 4x - 3x - 6 < 0\)
\(2x(x + 2) - 3(x + 2) < 0 \implies (2x - 3)(x + 2) < 0\)
If \((2x - 3)(x + 2) < 0\), then either (2x - 3) < 0 or (x + 2) < 0.
\((2x - 3) < 0 \implies x < \frac{3}{2}\), then \((x + 2) > 0 \implies x > -2\)
\((x + 2) < 0 \implies x < -2\), then \((2x - 3) > 0 \implies x > \frac{3}{2}\)
Check: Let x = -3,
\((2x - 3)(x + 2) = (-9)(-1) = 9 > 0\)
Let x = 1,
\((2x - 3)(x + 2) = (-1)(3) = -3 < 0\)
\(\therefore -2 < x < \frac{3}{2}\).
(b) \(\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}}\)
= \((\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}})(\frac{3\sqrt{5} - \sqrt{2}}{3\sqrt{5} - \sqrt{2}})\)
= \(\frac{15\sqrt{5} - 5\sqrt{2} - 30\sqrt{2} + 4\sqrt{5}}{9(5) - 3\sqrt{10} + 3\sqrt{10} - 2}\)
= \(\frac{19\sqrt{5} - 35\sqrt{2}}{45 - 2}\)
= \(\frac{19\sqrt{5} - 35\sqrt{2}}{43}\)
= \(\frac{-35}{43} \sqrt{2} + \frac{19}{43} \sqrt{5}\)
\(\therefore m = -\frac{-35}{43} ; n = \frac{19}{43}\).