I got 1/6 for (a) but (b) is beyond me. Thanks.
1/6 is right.
For b, you just want to show that the left is equal to the right. We do this by substituting and simplifying as follows:
$\displaystyle (r-p)^2=9-6\sqrt{2}$
$\displaystyle (\sqrt{6}-\sqrt{3})^2 \Rightarrow (\sqrt{6})^2-2\sqrt{6}\sqrt{3}+(\sqrt{3})^2 \Rightarrow 6-2\cdot 3\sqrt{2}+3 \Rightarrow 9-6\sqrt{2}$ which was to be shown.
$\displaystyle (\sqrt{6} - \sqrt{3})(\sqrt{6} - \sqrt{3})$
$\displaystyle = (\sqrt{6})^2 - \sqrt{18} - \sqrt{18} + \sqrt{9}$
$\displaystyle = 6 - 2\sqrt{18} + 3$
$\displaystyle = 9 - 2\sqrt{18}$
$\displaystyle = 9 - 2\sqrt{9 \cdot 2}$
$\displaystyle = 9 - 2\sqrt{3^2 \cdot 2}$
$\displaystyle = 9 - 2 \cdot3 \sqrt{2}$
$\displaystyle = 9 - 6\sqrt{2}$
-Dan