If \(f(x)=\begin{cases} \frac{1-\sin ^{5}x}{5\cos ^{2}x} & ,x\lt \frac{\pi }{2} \\ a & ,\text{at }x=\frac{\pi }{2} \\ \frac{b(1-\sin x)}{8(\pi -2x)^{2}} & ,x>\frac{\pi }{2} \end{cases}\) is continuous at \(x=\frac{\pi }{2}\), then \(\left(\frac{b}{a}\right)^{\frac{5}{3}}=\ldots\)