(1) Express \( \frac{\sqrt{3}-3 \sqrt{2}}{\sqrt{3}+3 \sqrt{2}} \) in the form \( a+b \sqrt{c} \)

To express \( \frac{\sqrt{3}-3 \sqrt{2}}{\sqrt{3}+3 \sqrt{2}} \) in the form \( a+b \sqrt{c} \), we can rationalize the denominator. We do this by multiplying both the numerator and the denominator by the conjugate of the denominator, which is \( \sqrt{3} - 3\sqrt{2} \). Calculating the numerator: \[ (\sqrt{3} - 3\sqrt{2})(\sqrt{3} - 3\sqrt{2}) = (\sqrt{3})^2 - 2(\sqrt{3})(3\sqrt{2}) + (3\sqrt{2})^2 = 3 - 6\sqrt{6} + 18 = 21 - 6\sqrt{6} \] Calculating the denominator: \[ (\sqrt{3} + 3\sqrt{2})(\sqrt{3} - 3\sqrt{2}) = (\sqrt{3})^2 - (3\sqrt{2})^2 = 3 - 18 = -15 \] Now we have: \[ \frac{\sqrt{3}-3\sqrt{2}}{\sqrt{3}+3\sqrt{2}} = \frac{21 - 6\sqrt{6}}{-15} \] We can separate this into two parts: \[ = -\frac{21}{15} + \frac{6\sqrt{6}}{15} \] Simplifying the fractions gives: \[ = -\frac{7}{5} + \frac{2\sqrt{6}}{5} \] Thus, we express it in the form \( a + b\sqrt{c} \): \[ = -\frac{7}{5} + \frac{2}{5}\sqrt{6} \] So, the final answer is: \[ -\frac{7}{5} + \frac{2}{5}\sqrt{6} \]