Step-by-step Solution:

1. Identify the Expression to Evaluate Let the given expression be denoted by $E$. We have: $E = 2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x - \int\limits_0^{\frac{\pi}{2}} g(x) d x$ The given functions are $f(x)=\sin^2 x$ and $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$.

2. Define the Integrals To simplify the problem, let's define two separate integrals:

   • $I_1 = \int\limits_0^{\frac{\pi}{2}} g(x) d x = \int\limits_0^{\frac{\pi}{2}} \sqrt{\frac{\pi x}{2}-x^2} d x$
   • $I_2 = \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x = \int\limits_0^{\frac{\pi}{2}} \sin^2(x) \sqrt{\frac{\pi x}{2}-x^2} d x$

   The expression to evaluate can now be written as $E = 2I_2 - I_1$.

3. Apply King's Property of Definite Integrals to $I_2$ We use the property $\int_0^a h(x) dx = \int_0^a h(a-x) dx$. In our case, $a = \frac{\pi}{2}$. Applying this property to the integral $I_2$: $I_2 = \int\limits_0^{\frac{\pi}{2}} \sin^2\left(\frac{\pi}{2}-x\right) \sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-x\right) - \left(\frac{\pi}{2}-x\right)^2} dx$

4. Simplify the Integrand Let's simplify the terms in the new integrand:

   • The trigonometric part: $\sin^2\left(\frac{\pi}{2}-x\right) = \cos^2(x)$.
   • The part under the square root, which is $g\left(\frac{\pi}{2}-x\right)$: $\frac{\pi}{2}\left(\frac{\pi}{2}-x\right) - \left(\frac{\pi}{2}-x\right)^2 = \frac{\pi^2}{4} - \frac{\pi x}{2} - \left(\frac{\pi^2}{4} - \pi x + x^2\right)$ $= \frac{\pi^2}{4} - \frac{\pi x}{2} - \frac{\pi^2}{4} + \pi x - x^2 = \frac{\pi x}{2} - x^2$ So, $\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-x\right) - \left(\frac{\pi}{2}-x\right)^2} = \sqrt{\frac{\pi x}{2}-x^2} = g(x)$.

5. Obtain a Second Expression for $I_2$ Substituting the simplified terms back into the integral for $I_2$, we get: $I_2 = \int\limits_0^{\frac{\pi}{2}} \cos^2(x) \sqrt{\frac{\pi x}{2}-x^2} dx = \int\limits_0^{\frac{\pi}{2}} \cos^2(x) g(x) dx$

6. Combine the Two Expressions for $I_2$ We now have two different expressions for $I_2$: (i) $I_2 = \int\limits_0^{\frac{\pi}{2}} \sin^2(x) g(x) dx$ (ii) $I_2 = \int\limits_0^{\frac{\pi}{2}} \cos^2(x) g(x) dx$

   Adding these two equations gives: $2I_2 = \int\limits_0^{\frac{\pi}{2}} \sin^2(x) g(x) dx + \int\limits_0^{\frac{\pi}{2}} \cos^2(x) g(x) dx$ $2I_2 = \int\limits_0^{\frac{\pi}{2}} (\sin^2(x) + \cos^2(x)) g(x) dx$

7. Simplify and Relate to $I_1$ Using the fundamental trigonometric identity $\sin^2(x) + \cos^2(x) = 1$, we get: $2I_2 = \int\limits_0^{\frac{\pi}{2}} (1) \cdot g(x) dx = \int\limits_0^{\frac{\pi}{2}} g(x) dx$ By our initial definition, this is exactly $I_1$. Therefore, we have the relationship: $2I_2 = I_1$

8. Calculate the Final Value The expression we need to find is $E = 2I_2 - I_1$. Substituting the result from the previous step, $2I_2 = I_1$, we get: $E = I_1 - I_1 = 0$

Thus, the value of the given expression is 0.