(a) Given that \( \beta=2+\sqrt{8} \), express \( \beta+\frac{1}{\beta} \) in the form \( a+b \sqrt{2} \), where \( a, b \), and \( c \) a the values of \( a \) and \( b \).

To express \( \beta + \frac{1}{\beta} \) in the form \( a + b \sqrt{2} \), we start with the given value of \( \beta \): \[ \beta = 2 + \sqrt{8} = 2 + 2\sqrt{2} \] Next, we need to calculate \( \frac{1}{\beta} \). We can do this by rationalizing the denominator: \[ \frac{1}{\beta} = \frac{1}{2 + 2\sqrt{2}} \cdot \frac{2 - 2\sqrt{2}}{2 - 2\sqrt{2}} = \frac{2 - 2\sqrt{2}}{(2 + 2\sqrt{2})(2 - 2\sqrt{2})} \] Now, we calculate the denominator: \[ (2 + 2\sqrt{2})(2 - 2\sqrt{2}) = 2^2 - (2\sqrt{2})^2 = 4 - 8 = -4 \] Thus, we have: \[ \frac{1}{\beta} = \frac{2 - 2\sqrt{2}}{-4} = -\frac{1}{2} + \frac{1}{2}\sqrt{2} \] Now we can find \( \beta + \frac{1}{\beta} \): \[ \beta + \frac{1}{\beta} = \left(2 + 2\sqrt{2}\right) + \left(-\frac{1}{2} + \frac{1}{2}\sqrt{2}\right) \] Combining the terms: 1. For the constant terms: \[ 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \] 2. For the terms involving \( \sqrt{2} \): \[ 2\sqrt{2} + \frac{1}{2}\sqrt{2} = \left(2 + \frac{1}{2}\right)\sqrt{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}\sqrt{2} \] Putting it all together, we have: \[ \beta + \frac{1}{\beta} = \frac{3}{2} + \frac{5}{2}\sqrt{2} \] Thus, in the form \( a + b\sqrt{2} \), we identify: \[ a = \frac{3}{2}, \quad b = \frac{5}{2} \] Therefore, the values of \( a \) and \( b \) are: \[ \boxed{\left(\frac{3}{2}, \frac{5}{2}\right)} \]