Let's simplify each of these expressions by rationalizing the denominator.

6. $\frac{1}{\sqrt{6} - 2}$

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, $\sqrt{6} + 2$:

$\frac{1}{\sqrt{6} - 2} \times \frac{\sqrt{6} + 2}{\sqrt{6} + 2} = \frac{\sqrt{6} + 2}{(\sqrt{6})^2 - 2^2} = \frac{\sqrt{6} + 2}{6 - 4} = \frac{\sqrt{6} + 2}{2} = \frac{\sqrt{6}}{2} + 1$

7. $\frac{7}{\sqrt{18} + 3}$

Rationalize by multiplying by the conjugate $\sqrt{18} - 3$:

$\frac{7}{\sqrt{18} + 3} \times \frac{\sqrt{18} - 3}{\sqrt{18} - 3} = \frac{7(\sqrt{18} - 3)}{(\sqrt{18})^2 - 3^2} = \frac{7(\sqrt{18} - 3)}{18 - 9} = \frac{7(\sqrt{18} - 3)}{9} = \frac{7\sqrt{18}}{9} - \frac{21}{9} = \frac{7\sqrt{18}}{9} - \frac{7}{3}$

Since $\sqrt{18} = 3\sqrt{2}$:

$\frac{7\sqrt{18}}{9} = \frac{7 \times 3\sqrt{2}}{9} = \frac{7\sqrt{2}}{3}$

Final result:

$\frac{7\sqrt{2}}{3} - \frac{7}{3} = \frac{7(\sqrt{2} - 1)}{3}$

8. $\frac{-12}{\sqrt{5} + 1}$

Rationalize by multiplying by the conjugate $\sqrt{5} - 1$:

[ \frac{-12}{\sqrt{5} + 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{-12