Step 1: Evaluate the limit to find a differential equation.

The given limit is: $L = \mathop {\lim }\limits_{t \to x} {{f(x)\sin t - f(t)\sin x} \over {t - x}}$ As $t \to x$, the numerator becomes $f(x)\sin x - f(x)\sin x = 0$ and the denominator becomes $x-x=0$. This is a $\frac{0}{0}$ indeterminate form, so we can apply L'Hôpital's rule by differentiating the numerator and denominator with respect to $t$.

$L = \mathop {\lim }\limits_{t \to x} {{\frac{d}{{dt}}(f(x)\sin t - f(t)\sin x)} \over {\frac{d}{{dt}}(t - x)}} = \mathop {\lim }\limits_{t \to x} {{f(x)\cos t - f'(t)\sin x} \over 1}$ Substituting $t=x$, we get: $L = f(x)\cos x - f'(x)\sin x$ We are given that $L = \sin^2x$. Therefore, we have the differential equation: $f(x)\cos x - f'(x)\sin x = \sin^2x$ Rearranging the terms, we get: $f'(x)\sin x - f(x)\cos x = -\sin^2x$

Step 2: Solve the differential equation.

For $x \in (0, \pi)$, $\sin x \neq 0$. We can divide the equation by $\sin^2x$: ${{f'(x)\sin x - f(x)\cos x} \over {\sin^2x}} = -1$ The left-hand side is the derivative of a quotient, specifically $\frac{d}{dx}\left(\frac{f(x)}{\sin x}\right)$. So, the equation becomes: ${d \over {dx}}\left( {{{f(x)} \over {\sin x}}} \right) = -1$ Integrating both sides with respect to $x$: $\int {{d \over {dx}}\left( {{{f(x)} \over {\sin x}}} \right)dx} = \int { - 1} dx$ ${{f(x)} \over {\sin x}} = -x + C$ where C is the constant of integration. Thus, the general solution is: $f(x) = (C-x)\sin x$ We use the given condition $f(\pi/6) = -\pi/12$ to find C. $f\left( {{\pi \over 6}} \right) = \left( {C - {\pi \over 6}} \right)\sin \left( {{\pi \over 6}} \right) = \left( {C - {\pi \over 6}} \right){1 \over 2}$ $-{\pi \over {12}} = {1 \over 2}\left( {C - {\pi \over 6}} \right)$ $-{\pi \over 6} = C - {\pi \over 6}$ This gives $C=0$. Therefore, the function is: $f(x) = -x\sin x$

Step 3: Evaluate each option.

Option A: $f\left( {{\pi \over 4}} \right) = {\pi \over {4\sqrt 2 }}$ Let's calculate $f(\pi/4)$ using our function: $f\left( {{\pi \over 4}} \right) = -\frac{\pi}{4}\sin\left(\frac{\pi}{4}\right) = -\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = -\frac{\pi}{4\sqrt{2}}$ The calculated value does not match the value given in the option. So, option A is FALSE.

Option B: $f(x) \lt {{{x^4}} \over 6} - {x^2}$ for all x $\in$(0, $\pi$) Substituting $f(x) = -x\sin x$, the inequality becomes: $-x\sin x < \frac{x^4}{6} - x^2$ $x^2 - x\sin x < \frac{x^4}{6}$ Since $x \in (0, \pi)$, $x>0$, so we can divide by $x$: $x - \sin x < \frac{x^3}{6}$ Let's define a function $g(x) = \frac{x^3}{6} - (x - \sin x)$. We want to show $g(x) > 0$ for $x \in (0, \pi)$. Let's analyze its derivatives: $g(0) = 0 - (0-0) = 0$. $g'(x) = \frac{3x^2}{6} - (1 - \cos x) = \frac{x^2}{2} - 1 + \cos x$. So, $g'(0) = 0 - 1 + 1 = 0$. $g''(x) = x - \sin x$. For $x > 0$, it is a known result that $x > \sin x$. Thus, $g''(x) > 0$ for $x \in (0, \pi)$. Since $g''(x) > 0$, $g'(x)$ is a strictly increasing function. For $x > 0$, $g'(x) > g'(0) = 0$. Since $g'(x) > 0$, $g(x)$ is also a strictly increasing function. For $x > 0$, $g(x) > g(0) = 0$. Therefore, the inequality $x - \sin x < \frac{x^3}{6}$ holds for $x \in (0, \pi)$. So, option B is TRUE.

Option C: There exists $\alpha \in$(0, $\pi$) such that f'($\alpha$) = 0 We have the function $f(x) = -x\sin x$. This function is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$. Let's evaluate the function at the endpoints of the interval $[0, \pi]$: $f(0) = -0 \cdot \sin(0) = 0$. $f(\pi) = -\pi \cdot \sin(\pi) = 0$. Since $f(0) = f(\pi)$, by Rolle's Theorem, there must exist at least one point $\alpha \in (0, \pi)$ such that $f'(\alpha)=0$. So, option C is TRUE.

Option D: $f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = 0$ First, find the first and second derivatives of $f(x) = -x\sin x$. $f'(x) = -(1 \cdot \sin x + x \cdot \cos x) = -\sin x - x\cos x$. $f''(x) = -\cos x - (1 \cdot \cos x + x(-\sin x)) = -\cos x - \cos x + x\sin x = x\sin x - 2\cos x$. Now, evaluate $f(\pi/2)$ and $f''(\pi/2)$: $f\left( {{\pi \over 2}} \right) = -{{\pi \over 2}}\sin\left({{\pi \over 2}}\right) = -{{\pi \over 2}}(1) = -{{\pi \over 2}}$ $f''\left( {{\pi \over 2}} \right) = {{\pi \over 2}}\sin\left({{\pi \over 2}}\right) - 2\cos\left({{\pi \over 2}}\right) = {{\pi \over 2}}(1) - 2(0) = {{\pi \over 2}}$ Now, check the condition: $f''\left( {{\pi \over 2}} \right) + f\left( {{\pi \over 2}} \right) = {{\pi \over 2}} + \left(-{{\pi \over 2}}\right) = 0$ The condition is satisfied. So, option D is TRUE.

Final conclusion: Options B, C, and D are true.