If `f(x)={(((1-sin^(3)x))/(3cos^(2)x)",",x lt (pi)/(2)),(a",",x=(pi)/(2)),((b(1-sinx))/((pi-2x)^(2))",",x gt (pi)/(2)):}`
is continuous at `x=(pi)/(2)`, then the value of `((b)/(a))^(5//3)` is

Explanation

To solve the problem, we need to ensure the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). The function is defined piecewise as follows:

\[ f(x) = \begin{cases} \frac{1 - \sin^3 x}{3 \cos^2 x} & \text{if } x < \frac{\pi}{2} \\ a & \text{if } x = \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2} & \text{if } x > \frac{\pi}{2} \end{cases} \]

### Step 1: Find \( f\left(\frac{\pi}{2}\right) \)

Since \( f\left(\frac{\pi}{2}\right) = a \), we need to find the left-hand limit \( \lim_{x \to \frac{\pi}{2}^-} f(x) \) and the right-hand limit \( \lim_{x \to \frac{\pi}{2}^+} f(x) \).

### Step 2: Calculate the left-hand limit

For \( x < \frac{\pi}{2} \):

\[ f(x) = \frac{1 - \sin^3 x}{3 \cos^2 x} \]

We need to evaluate:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^3 x}{3 \cos^2 x} \]

As \( x \to \frac{\pi}{2} \):
- \( \sin x \to 1 \) so \( \sin^3 x \to 1 \)
- \( \cos x \to 0 \) so \( \cos^2 x \to 0 \)

Thus, we have:

\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \frac{1 - 1}{3 \cdot 0} = \frac{0}{0} \]

This is an indeterminate form, so we apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator:

1. **Numerator**: \( \frac{d}{dx}(1 - \sin^3 x) = -3 \sin^2 x \cos x \)
2. **Denominator**: \( \frac{d}{dx}(3 \cos^2 x) = -6 \cos x \sin x \)

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{2}^-} \frac{-3 \sin^2 x \cos x}{-6 \cos x \sin x} = \lim_{x \to \frac{\pi}{2}^-} \frac{3 \sin x}{2} = \frac{3 \cdot 1}{2} = \frac{3}{2} \]

### Step 4: Calculate the right-hand limit

For \( x > \frac{\pi}{2} \):

\[ f(x) = \frac{b(1 - \sin x)}{(\pi - 2x)^2} \]

We need to evaluate:

\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{b(1 - \sin x)}{(\pi - 2x)^2} \]

As \( x \to \frac{\pi}{2} \):
- \( \sin x \to 1 \) so \( 1 - \sin x \to 0 \)
- \( \pi - 2x \to 0 \)

This also results in an indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 5: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:

1. **Numerator**: \( \frac{d}{dx}(b(1 - \sin x)) = -b \cos x \)
2. **Denominator**: \( \frac{d}{dx}((\pi - 2x)^2) = -4(\pi - 2x) \)

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{-b \cos x}{-4(\pi - 2x)} = \lim_{x \to \frac{\pi}{2}^+} \frac{b \cos x}{4(\pi - 2x)} \]

As \( x \to \frac{\pi}{2} \):
- \( \cos x \to 0 \)
- \( \pi - 2x \to 0 \)

This again results in \( \frac{0}{0} \), so we apply L'Hôpital's Rule a second time.

### Step 6: Apply L'Hôpital's Rule again

Differentiate again:

1. **Numerator**: \( \frac{d}{dx}(b \cos x) = -b \sin x \)
2. **Denominator**: \( \frac{d}{dx}(-4(\pi - 2x)) = 8 \)

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{2}^+} \frac{-b \sin x}{8} = \frac{-b \cdot 1}{8} = -\frac{b}{8} \]

### Step 7: Set the limits equal for continuity

For continuity at \( x = \frac{\pi}{2} \):

\[ \frac{3}{2} = a = -\frac{b}{8} \]

From \( a = \frac{3}{2} \), we have:

\[ b = -8a = -8 \cdot \frac{3}{2} = -12 \]

### Step 8: Find \( \frac{b}{a} \)

Now we calculate \( \frac{b}{a} \):

\[ \frac{b}{a} = \frac{-12}{\frac{3}{2}} = -12 \cdot \frac{2}{3} = -8 \]

### Step 9: Calculate \( \left(\frac{b}{a}\right)^{\frac{5}{3}} \)

Finally, we find:

\[ \left(\frac{b}{a}\right)^{\frac{5}{3}} = (-8)^{\frac{5}{3}} = -32 \]

Thus, the final answer is:

\[ \boxed{-32} \]