The user wants me to solve a multiple-choice question with multiple correct answers from the topic of calculus.

Given Information:

1. $f(x)$ is a non-constant, twice differentiable function on $(-\infty, \infty)$.
2. $f(x) = f(1-x)$ for all $x$. This implies the function is symmetric about the line $x = 1/2$.
3. $f'(1/4) = 0$.

Let's analyze the properties of $f(x)$ and then evaluate each option.

Analysis of $f(x) = f(1-x)$

Differentiating the given equation with respect to $x$, we get: $f'(x) = f'(1-x) \cdot \frac{d}{dx}(1-x)$ $f'(x) = -f'(1-x) \quad (*)$ Differentiating again with respect to $x$: $f''(x) = -f''(1-x) \cdot (-1)$ $f''(x) = f''(1-x) \quad (**)$ This shows that $f'(x)$ is anti-symmetric about $x=1/2$, and $f''(x)$ is symmetric about $x=1/2$.

Evaluation of Option B: $f'\left( {{1 \over 2}} \right) = 0$

1. Using the relation $f'(x) = -f'(1-x)$ from equation $(*)$.
2. Substitute $x = 1/2$ into the equation: $f'(1/2) = -f'(1 - 1/2)$ $f'(1/2) = -f'(1/2)$
3. Rearranging the terms, we get: $2f'(1/2) = 0$ $f'(1/2) = 0$

Conclusion: Option B is correct.

Evaluation of Option A: $f''\left( x \right)$ vanishes at least twice on $\left[ {0,1} \right]$

1. We are given that $f'(1/4) = 0$.
2. From equation $(*)$, let's find the value of $f'(3/4)$: $f'(3/4) = -f'(1 - 3/4) = -f'(1/4)$.
3. Since $f'(1/4) = 0$, it follows that $f'(3/4) = 0$.
4. We have now established three points where the first derivative is zero: $x = 1/4$, $x = 1/2$ (from option B), and $x = 3/4$. All these points lie in the interval $[0, 1]$.
5. Consider the function $g(x) = f'(x)$. Since $f(x)$ is twice differentiable, $f'(x)$ is continuous and differentiable everywhere.
6. Apply Rolle's Theorem to $f'(x)$ on the interval $[1/4, 1/2]$:
   • $f'(x)$ is continuous on $[1/4, 1/2]$.
   • $f'(x)$ is differentiable on $(1/4, 1/2)$.
   • $f'(1/4) = f'(1/2) = 0$. Therefore, there exists at least one point $c_1 \in (1/4, 1/2)$ such that $f''(c_1) = 0$.
7. Apply Rolle's Theorem to $f'(x)$ on the interval $[1/2, 3/4]$:
   • $f'(x)$ is continuous on $[1/2, 3/4]$.
   • $f'(x)$ is differentiable on $(1/2, 3/4)$.
   • $f'(1/2) = f'(3/4) = 0$. Therefore, there exists at least one point $c_2 \in (1/2, 3/4)$ such that $f''(c_2) = 0$.
8. Since $c_1 < 1/2$ and $c_2 > 1/2$, the points $c_1$ and $c_2$ are distinct. Both points lie within the interval $[0, 1]$.

Conclusion: Option A is correct.

Evaluation of Option C: $\int\limits_{ - 1/2}^{1/2} {f\left( {x + {1 \over 2}} \right)\sin x\,dx} = 0$

1. Let the integral be $I = \int_{-1/2}^{1/2} f(x + 1/2) \sin x \,dx$.
2. We check if the integrand, $g(x) = f(x + 1/2) \sin x$, is an odd function. An integral of an odd function over a symmetric interval $[-a, a]$ is zero.
3. Evaluate $g(-x)$: $g(-x) = f(-x + 1/2) \sin(-x) = f(1/2 - x) (-\sin x) = -f(1/2 - x) \sin x$.
4. The given property $f(t) = f(1-t)$ implies symmetry about $t=1/2$. Let's set $t = 1/2 + x$. Then $1-t = 1 - (1/2 + x) = 1/2 - x$. So, $f(1/2 + x) = f(1/2 - x)$.
5. Substitute this back into the expression for $g(-x)$: $g(-x) = -f(1/2 + x) \sin x = -g(x)$.
6. Since $g(x)$ is an odd function, its integral over the symmetric interval $[-1/2, 1/2]$ is zero.

Conclusion: Option C is correct.

Evaluation of Option D: $\int\limits_0^{1/2} {f\left( t \right){e^{\sin \,\pi t}}dt = } \int\limits_{1/2}^1 {f\left( {1 - t} \right){e^{\sin \,\pi t}}dt}$

1. Let's analyze the right-hand side (RHS) of the equation: $RHS = \int_{1/2}^1 f(1-t) e^{\sin(\pi t)} dt$.
2. We use a substitution. Let $u = 1-t$. Then $t = 1-u$ and $dt = -du$.
3. We also need to change the limits of integration:
   • When $t = 1/2$, $u = 1 - 1/2 = 1/2$.
   • When $t = 1$, $u = 1 - 1 = 0$.
4. Substituting these into the RHS integral: $RHS = \int_{1/2}^{0} f(u) e^{\sin(\pi(1-u))} (-du) = \int_{0}^{1/2} f(u) e^{\sin(\pi - \pi u)} du$.
5. Using the trigonometric identity $\sin(\pi - \theta) = \sin(\theta)$, we get: $RHS = \int_{0}^{1/2} f(u) e^{\sin(\pi u)} du$.
6. This expression is identical to the left-hand side (LHS), with the variable of integration being $u$ instead of $t$. The name of the integration variable does not affect the value of the definite integral. $RHS = \int_{0}^{1/2} f(t) e^{\sin(\pi t)} dt = LHS$.
7. The equality holds. Note that this derivation is valid for any integrable function $f$ and does not depend on the specific properties of $f$ given in the problem, such as $f(x)=f(1-x)$. It is a mathematical identity.

Conclusion: Option D is correct.

Final Summary

Based on the step-by-step analysis, all four options A, B, C, and D are correct statements derived from the given information or are general mathematical identities. The question asks for which statements follow from the premises. All four statements are true.

• A: Correct, by applying Rolle's theorem on $f'(x)$.
• B: Correct, a direct consequence of $f'(x)=-f'(1-x)$.
• C: Correct, based on the integrand being an odd function.
• D: Correct, as it is a mathematical identity proven by substitution.

Therefore, the correct answer should include all four options. The stored answer is {A, B, C}, which is incomplete.