1. We are given

$$f(t)=\int_0^{\pi} \frac{2x\,dx}{1-\cos^2 t\,\sin^2 x}, \qquad 0<t<\pi$$

and we need to find

$$I=\int_0^{\pi/2} \frac{\pi^2\,dt}{f(t)}.$$

2. First, simplify $f(t)$ using symmetry.

Write

$$J=\int_0^{\pi} \frac{2x\,dx}{1-a\sin^2 x},$$

where

$$a=\cos^2 t.$$

Now use the substitution $x\mapsto \pi-x$:

$$J=\int_0^{\pi} \frac{2(\pi-x)\,dx}{1-a\sin^2 x},$$

because $\sin(\pi-x)=\sin x$.

Adding the two expressions for $J$,

$$2J=\int_0^{\pi} \frac{2x+2(\pi-x)}{1-a\sin^2 x}\,dx =\int_0^{\pi} \frac{2\pi\,dx}{1-a\sin^2 x}.$$

Hence

$$J=\pi\int_0^{\pi} \frac{dx}{1-a\sin^2 x}.$$

Therefore,

$$f(t)=\pi\int_0^{\pi} \frac{dx}{1-\cos^2 t\sin^2 x}.$$

3. Evaluate the standard integral.

Using symmetry again,

$$\int_0^{\pi} \frac{dx}{1-a\sin^2 x}=2\int_0^{\pi/2} \frac{dx}{1-a\sin^2 x}.$$

Now the standard result is

$$\int_0^{\pi/2} \frac{dx}{1-a\sin^2 x}=\frac{\pi}{2\sqrt{1-a}}, \qquad 0\le a<1.$$

So,

$$\int_0^{\pi} \frac{dx}{1-a\sin^2 x}=\frac{\pi}{\sqrt{1-a}}.$$

With $a=\cos^2 t$,

$$\sqrt{1-a}=\sqrt{1-\cos^2 t}=\sin t$$

for $0<t<\pi$. Thus,

$$f(t)=\pi\cdot \frac{\pi}{\sin t}=\frac{\pi^2}{\sin t}.$$

4. Now compute the required integral:

$$I=\int_0^{\pi/2} \frac{\pi^2}{f(t)}\,dt =\int_0^{\pi/2} \frac{\pi^2}{\pi^2/\sin t}\,dt =\int_0^{\pi/2} \sin t\,dt.$$

Therefore,

$$I=[-\cos t]_0^{\pi/2}=0-(-1)=1.$$

5. Final answer:

$$\boxed{1}$$

The derived answer matches the stored correct answer.