Now, I know the basic steps of conjugating a number; it's basically a difference of squares.
So to solve both #'s 1 & 2, I did the following:
1.) $\frac {7{\sqrt 2}}{\sqrt{6} + 8 }\cdot\frac {\sqrt {6} - 8}{\sqrt {6} - 8}$ = $\frac {- 56{\sqrt 12}}{6 - 64 }$ = $\frac {- 56{\sqrt3\cdot4}}{- 58 }$ = $\frac {- 112{\sqrt 3}}{- 58 }$
2.) $\frac {{\sqrt 3} + {\sqrt 13}}{\sqrt{3} - \sqrt {13} }\cdot\frac {{\sqrt 3} + {\sqrt 13}}{\sqrt{3} + \sqrt {13} }$ = $\frac {3 + 13}{3 - 13 }$ = $-\frac {16}{10}$
and got $\frac {- 112{\sqrt 3}}{- 58 }$ for #1
and $-\frac {16}{10}$ for #2
The answers in the textbook are:
1.) $\frac {- 7{\sqrt 3} + 28\sqrt {2}}{29 }$
2.) $\frac {- 8 -\sqrt {39}}{5 }$
Possibly the product from the first step is wrong, but I don't know what mistake I made.
You failed to write down all the terms while expanding the numerator after conjugation. For (1): $$\frac{7\sqrt2}{\sqrt6+8}\cdot\frac{\sqrt6-8}{\sqrt6-8}=\frac{7\sqrt{12}-56\sqrt2}{6-64}=\frac{14\sqrt3-56\sqrt2}{-58}=\frac{28\sqrt2-7\sqrt3}{29}$$ For (2): $$\frac{\sqrt3+\sqrt{13}}{\sqrt3-\sqrt{13}}\cdot\frac{\sqrt3+\sqrt{13}}{\sqrt3+\sqrt{13}}=\frac{3+2\sqrt{39}+13}{3-13}=\frac{16+2\sqrt{39}}{-10}=\frac{-8-\sqrt{39}}5$$