For the first part, the results get quite crazy. I hardly believe that you're supposed to do this by hand, so I'm afraid that there's some typo in the question.
As for the second part, we have
[tex]\sqrt{x}+\dfrac{1}{\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}[/tex]
Plug your value for x and you have
[tex]\dfrac{x+1}{\sqrt{x}} \mapsto \dfrac{4+2\sqrt{2}}{\sqrt{3+2\sqrt{2}}}=\dfrac{4+2\sqrt{2}}{1+\sqrt{2}}[/tex]
We can rewrite the quantity as
[tex]\dfrac{4+2\sqrt{2}}{1+\sqrt{2}}\cdot\dfrac{1-\sqrt{2}}{1-\sqrt{2}}=\dfrac{4-4\sqrt{2}+2\sqrt{2}-4}{-1} = 2\sqrt{2}[/tex]
