(a) \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\) \(1\frac{1}{4} + \frac{7}{9} = \frac{5}{4} + \frac{7}{9}\) = \(\frac{45 + 28}{36}\) = \(\frac{73}{36}\) \(1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64} = \frac{13}{9} - (\frac{8}{3} \times \frac{9}{64})\) = \(\frac{13}{9} - \frac{3}{8}\) = \(\frac{104 - 27}{72}\) = \(\frac{77}{72}\) \(\therefore \frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}} = \frac{73}{36} \div \frac{77}{72}\) = \(\frac{73}{36} \times \frac{72}{77}\) = \(\frac{146}{77}\) = \(1\frac{69}{77}\) (b) \(\sin x = \frac{2}{3}\) In \(\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\) \(\therefore |BC|^{2} = 9 - 4 = 5\) \(|BC| = \sqrt{5}\) \(\tan x = \frac{2}{\sqrt{5}} ; \cos x = \frac{\sqrt{5}}{3}\) \(\therefore \tan x - \cos x = \frac{2}{\sqrt{5}} - \frac{\sqrt{5}}{3}\) = \(\frac{6 - 5}{3\sqrt{5}}\) = \(\frac{1}{3\sqrt{5}}\) Rationalizing, we have \(\frac{1}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{15}\)