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 that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \). The function is defined piecewise, and we will evaluate the limits from both sides of \( x = \frac{\pi}{2} \).

### Step 1: Evaluate \( \lim_{x \to \frac{\pi}{2}^-} f(x) \)

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

We need to find the limit as \( x \) approaches \( \frac{\pi}{2} \):
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin^3 x}{3 \cos^2 x} \]

Substituting \( x = \frac{\pi}{2} \):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \). We will apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:
- The derivative of the numerator \( 1 - \sin^3 x \) is \( -3 \sin^2 x \cos x \).
- The derivative of the denominator \( 3 \cos^2 x \) is \( -6 \cos x \sin x \).

Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{-3 \sin^2 x \cos x}{-6 \cos x \sin x} = \lim_{x \to \frac{\pi}{2}} \frac{3 \sin x}{2} = \frac{3 \cdot 1}{2} = \frac{3}{2} \]

### Step 3: Set \( f\left(\frac{\pi}{2}\right) = a \)

Since the function is continuous at \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = a \]
Thus, we have:
\[ a = \frac{3}{2} \]

### Step 4: Evaluate \( \lim_{x \to \frac{\pi}{2}^+} f(x) \)

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

We need to find the limit as \( x \) approaches \( \frac{\pi}{2} \):
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{b(1 - \sin x)}{(\pi - 2x)^2} \]

Substituting \( x = \frac{\pi}{2} \):
\[ 1 - \sin\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \pi - 2\left(\frac{\pi}{2}\right) = 0 \]
This again gives us the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 5: Apply L'Hôpital's Rule Again

Differentiate the numerator and denominator:
- The derivative of the numerator \( b(1 - \sin x) \) is \( -b \cos x \).
- The derivative of the denominator \( (\pi - 2x)^2 \) is \( -4(\pi - 2x) \).

Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}} \frac{-b \cos x}{-4(\pi - 2x)} = \lim_{x \to \frac{\pi}{2}} \frac{b \cos x}{4(\pi - 2x)} \]
Substituting \( x = \frac{\pi}{2} \):
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \pi - 2\left(\frac{\pi}{2}\right) = 0 \]
We apply L'Hôpital's Rule again.

### Step 6: Differentiate Again

Differentiate again:
- The derivative of the numerator \( b \cos x \) is \( -b \sin x \).
- The derivative of the denominator \( 4(\pi - 2x) \) is \( -8 \).

Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to \frac{\pi}{2}} \frac{-b \sin x}{-8} = \lim_{x \to \frac{\pi}{2}} \frac{b \sin x}{8} = \frac{b \cdot 1}{8} = \frac{b}{8} \]

### Step 7: Set Limits Equal for Continuity

Setting the limits equal for continuity:
\[ \frac{b}{8} = a = \frac{3}{2} \]
Thus,
\[ b = 8 \cdot \frac{3}{2} = 12 \]

### Step 8: Calculate \( \left(\frac{b}{a}\right)^{\frac{5}{3}} \)

Now we can find \( \frac{b}{a} \):
\[ \frac{b}{a} = \frac{12}{\frac{3}{2}} = 12 \cdot \frac{2}{3} = 8 \]

Finally, we compute:
\[ \left(\frac{b}{a}\right)^{\frac{5}{3}} = 8^{\frac{5}{3}} = (2^3)^{\frac{5}{3}} = 2^5 = 32 \]

### Final Answer
The value of \( \left(\frac{b}{a}\right)^{\frac{5}{3}} \) is \( \boxed{32} \).

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