a) Denote the arithmetic mean given in the problem by $A(x)$. As
$$1 + \sin x = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right)^2,$$

$$1 - \sin x = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left( \sin \frac{x}{2} - \cos \frac{x}{2} \right)^2,$$
we get
$$A(x) = \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{2} = \frac{|\sin \frac{x}{2} + \cos \frac{x}{2}| + |\sin \frac{x}{2} - \cos \frac{x}{2}|}{2}.$$
Depending on the signs of the numbers $\sin \frac{x}{2} + \cos \frac{x}{2}$ and $\sin \frac{x}{2} - \cos \frac{x}{2}$, one of the trigonometric functions in the numerator cancels out and the other one is doubled, with either a positive or a negative sign. Therefore, $A(x)$ is equal to one of the numbers $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, $-\sin \frac{x}{2}$, $-\cos \frac{x}{2}$.

b) Clearly $A(x) = 1$, whenever $x$ is one of the numbers $0, \pi, 2\pi, 3\pi$. Nevertheless, each of these four values makes a unique expression among $\sin \frac{x}{2}, \cos \frac{x}{2}, -\sin \frac{x}{2}, -\cos \frac{x}{2}$ evaluate to 1. Therefore, none of these four can be left out.

Part a) can also be proven as follows. Let $A(x)$ be the same as in the first solution. Then
$$\left( \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{2} \right)^2 = \frac{2 + 2\sqrt{1 - \sin^2 x}}{4} = \frac{1 + |\cos x|}{2},$$
so that $A(x) = \sqrt{\frac{1+|\cos x|}{2}}$. Therefore, if $\cos x \ge 0$, then $A(x) = \sqrt{\frac{1+\cos x}{2}} = \pm \cos \frac{x}{2}$; if $\cos x < 0$, then $A(x) = \sqrt{\frac{1-\cos x}{2}} = \pm \sin \frac{x}{2}$.