8 Given that \( \frac{2}{(1+\sqrt{2})+\sqrt{3}} \) can be expressed in the form \( \frac{1}{2}(\sqrt{a}-\sqrt{b}+c) \), find the value of \( a \), of \( b \) and of \( c \).

To simplify \( \frac{2}{(1+\sqrt{2})+\sqrt{3}} \), we first rewrite the denominator: \[ (1+\sqrt{2}) + \sqrt{3} = 1 + \sqrt{2} + \sqrt{3}. \] Now, let's simplify \( \frac{2}{1 + \sqrt{2} + \sqrt{3}} \) by rationalizing the denominator. We multiply the numerator and the denominator by the conjugate of the denominator: \[ \text{Conjugate: } (1 + \sqrt{2} - \sqrt{3}). \] Now, the expression becomes: \[ \frac{2(1 + \sqrt{2} - \sqrt{3})}{(1 + \sqrt{2} + \sqrt{3})(1 + \sqrt{2} - \sqrt{3})}. \] Next, we simplify the denominator: \[ (1 + \sqrt{2} + \sqrt{3})(1 + \sqrt{2} - \sqrt{3}) = (1 + \sqrt{2})^2 - (\sqrt{3})^2. \] Calculating \( (1 + \sqrt{2})^2 \): \[ (1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}. \] Now, the denominator becomes: \[ (3 + 2\sqrt{2}) - 3 = 2\sqrt{2}. \] Now substituting back into the expression, we have: \[ \frac{2(1 + \sqrt{2} - \sqrt{3})}{2\sqrt{2}} = \frac{1 + \sqrt{2} - \sqrt{3}}{\sqrt{2}}. \] Separating the terms gives: \[ \frac{1}{\sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2}} - \frac{\sqrt{3}}{\sqrt{2}} = \frac{1}{\sqrt{2}} + 1 - \frac{\sqrt{3}}{\sqrt{2}}. \] Combining and using the fact that \( \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \): \[ 1 + \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{\sqrt{2}}. \] Rationalizing \( \frac{\sqrt{3}}{\sqrt{2}} \) as \( \frac{\sqrt{6}}{2} \): Thus, \[ 1 + \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2} = 1 + \frac{1}{2}(\sqrt{2} - \sqrt{6}). \] Now, we need to express this in the form \( \frac{1}{2}(\sqrt{a} - \sqrt{b} + c) \). Matching terms, we find: \[ c = 2, \quad a = 6, \quad b = 2. \] Thus, the final answers are: \[ \boxed{6}, \quad \boxed{2}, \quad \boxed{2}. \]