(1) We have
$$f(α)= \frac {\sin(π-α)\cos(2π-α)\sin(-α+ \frac {3π}{2})}{\sin( \frac {π}{2}+\alpha )\sin(-π -\alpha )}.$$
Using the trigonometric identities, $\sin(π-α) = \sin α$, $\cos(2π-α) = \cos α$, $\sin(-α+ \frac {3π}{2}) = -\cos α$, and $\sin(\frac {π}{2}+\alpha) = \cos α$, $\sin(-π -\alpha ) = \sin α$, the expression simplifies to
$$f(α) = \frac {\sin α \cdot \cos α \cdot (-\cos α)}{\cos α \cdot \sin α} = -\cos α.$$

(2) Given that $α$ is an angle in the third quadrant, and $\cos\left(α+ \frac {π}{3}\right) = \frac {3}{5} > 0$, so $α+ \frac {π}{3}$ is an angle in the fourth quadrant.
Thus,
$$\sin\left(α+ \frac {π}{3}\right) = - \sqrt {1-\cos^{2}\left(α+ \frac {π}{3}\right)}= - \sqrt {1-\left(\frac {3}{5}\right)^2}= - \frac {4}{5}.$$
Now, we simplify $f(α)$:
$$\begin{align*} f(α) &= -\cos α \\ &= -\cos\left[\left(α+ \frac {π}{3}\right) - \frac {π}{3}\right] \\ &= -\cos\left(α+ \frac {π}{3}\right)\cos \frac {π}{3} + \sin\left(α+ \frac {π}{3}\right)\sin \frac {π}{3} \\ &= -\frac {3}{5} \cdot \frac {1}{2} + (-\frac {4}{5})\cdot \frac {\sqrt {3}}{2} \\ &= -\frac {3}{10} - \frac {4\sqrt {3}}{10} \\ &= \boxed{\frac {-3 - 4\sqrt {3}}{10}}. \end{align*}$$