Using the trigonometric identities and simplification, we have
$$f(\alpha) = \frac{\sin(2\pi - \alpha)\cos\left(\frac{\pi}{2} + \alpha\right)}{\cos\left(-\frac{\pi}{2} + \alpha\right)\tan(\pi + \alpha)}.$$
This simplifies to
$$f(\alpha) = \frac{-\sin(\alpha)(-\sin(\alpha))}{\sin(\alpha)\tan(\alpha)}.$$

Now, because $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$, we get
$$f(\alpha) = \frac{\sin^2(\alpha)}{\sin(\alpha)\cdot \frac{\sin(\alpha)}{\cos(\alpha)}} = \cos(\alpha).$$

Therefore, $f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \boxed{\frac{1}{2}}$.

Option A is correct.