Step-by-step Solution:

1. Analyze Statement 2: Statement 2 is: ${\sin ^2}(x + \pi ) = {\sin ^2}x$ for all real x.

We know the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$. Let A=x and $B=\pi$. $\sin(x + \pi) = \sin x \cos \pi + \cos x \sin \pi$ Since $\cos \pi = -1$ and $\sin \pi = 0$, we have: $\sin(x + \pi) = (\sin x)(-1) + (\cos x)(0) = -\sin x$

Now, we square both sides: ${\sin ^2}(x + \pi ) = ( - \sin x)^2 = {\sin ^2}x$ This identity is true for all real values of x. Therefore, Statement 2 is True.

2. Analyze Statement 1: Statement 1 is: The function F(x) satisfies $F(x+\pi) = F(x)$ for all real x, where F(x) is an indefinite integral of $\sin^2x$.

First, we need to find the indefinite integral F(x). $F(x) = \int {\sin ^2}x\,dx$ To integrate $\sin^2x$, we use the trigonometric identity $\cos(2x) = 1 - 2{\sin^2}x$, which can be rearranged to ${\sin^2}x = \frac{1 - \cos(2x)}{2}$.

Substituting this into the integral: $F(x) = \int {\frac{{1 - \cos(2x)}}{2}} \,dx$ $F(x) = \frac{1}{2}\int {(1 - \cos(2x))} \,dx$ $F(x) = \frac{1}{2}\left[ {x - \frac{{\sin(2x)}}{2}} \right] + C$ $F(x) = \frac{x}{2} - \frac{{\sin(2x)}}{4} + C$ where C is the constant of integration.

3. Check the condition $F(x+\pi) = F(x)$: Now, let's evaluate $F(x+\pi)$: $F(x + \pi ) = \frac{{x + \pi }}{2} - \frac{{\sin(2(x + \pi ))}}{4} + C$ $F(x + \pi ) = \frac{x}{2} + \frac{\pi }{2} - \frac{{\sin(2x + 2\pi )}}{4} + C$ We know that the sine function has a period of $2\pi$, so $\sin(2x + 2\pi) = \sin(2x)$. $F(x + \pi ) = \frac{x}{2} + \frac{\pi }{2} - \frac{{\sin(2x)}}{4} + C$ Rearranging the terms to compare with F(x): $F(x + \pi ) = \left( {\frac{x}{2} - \frac{{\sin(2x)}}{4} + C} \right) + \frac{\pi }{2}$ $F(x + \pi ) = F(x) + \frac{\pi }{2}$ Since $F(x + \pi) = F(x) + \frac{\pi}{2}$, and $\frac{\pi}{2} \neq 0$, the condition $F(x+\pi) = F(x)$ is not satisfied. Therefore, Statement 1 is False.

4. Conclusion: We have found that Statement 1 is False and Statement 2 is True. Let's check the given options:

• A: Statement 1 is True, ... (Incorrect)
• B: Statement 1 is True, ... (Incorrect)
• C: Statement 1 is True, ... (Incorrect)
• D: Statement 1 is False, Statement 2 is True. (Correct)

Alternative perspective: Statement 2 implies that the integrand $f(x) = \sin^2x$ is a periodic function with period $\pi$. For its indefinite integral F(x) to also be periodic with period $\pi$, the average value of f(x) over one period must be zero. That is, $\int_0^\pi f(x) dx = 0$. Let's calculate this definite integral: $\int_0^\pi {\sin ^2}x\,dx = \int_0^\pi {\frac{{1 - \cos(2x)}}{2}} \,dx = \frac{1}{2}\left[ {x - \frac{{\sin(2x)}}{2}} \right]_0^\pi$ $= \frac{1}{2}\left[ {(\pi - \frac{{\sin(2\pi )}}{2}) - (0 - \frac{{\sin(0)}}{2})} \right] = \frac{1}{2}[(\pi - 0) - (0 - 0)] = \frac{\pi }{2}$ Since $\int_0^\pi {\sin ^2}x\,dx = \frac{\pi }{2} \ne 0$, the function F(x) is not periodic with period $\pi$. This confirms that Statement 1 is False.