To find the ratio of the masses $$\frac{m_2}{m_1}$$ given that the acceleration of the system is $$\frac{g}{8}$$, where $$g$$ is the acceleration due to gravity, we need to analyze the forces acting on each mass and apply Newton's second law of motion.

For mass $$m_1$$, the force acting downward (towards the centre of the Earth) is the gravitational force $$m_1g$$, and the tension $$T$$ acts upward. Its equation of motion, considering downward as positive, can be given by:

$m_1g - T = m_1a$

For mass $$m_2$$, the force acting downward is $$m_2g$$, but since it is on the opposite side of the pulley, the tension $$T$$ is upwards (considering upward movement as positive direction), so we have:

$T - m_2g = m_2a$

Since the pulley is light and smooth, the tension $$T$$ is the same on both sides of the pulley. Also, the system accelerates together, so $$a = \frac{g}{8}$$.

Adding the two equations to eliminate $$T$$ gives us:

$m_1g - m_2g = (m_1 - m_2)a$

$g(m_1 - m_2) = (m_1 - m_2)\frac{g}{8}$

Canceling out $$g$$ and dividing both sides by $$m_1 - m_2$$ (assuming $$m_1 \neq m_2$$), we get:

$1 = \frac{1}{8}$

This simplification doesn't align with finding the ratio directly, indicating a mistake in handling the simultaneous equations relation to $$a$$ and $$T$$. Let's correct our approach to finding the ratio of the masses based on the acceleration and tension.

Since the acceleration $$a = \frac{g}{8}$$, and considering the system as a whole, the net force causing the acceleration is the difference in gravitational forces on the two masses. This net force provides the system's entire acceleration. Thus, correctly setting up the equations for the system should give:

$m_2g - m_1g = (m_1 + m_2)\frac{g}{8}$

$g(m_2-m_1) = \frac{g}{8}(m_1 + m_2)$

$8(m_2 - m_1) = m_1 + m_2$

$8m_2 - 8m_1 = m_1 + m_2$

$7m_2 = 9m_1$

$\frac{m_2}{m_1} = \frac{9}{7}$

Therefore, the correct ratio of the masses $$\frac{m_2}{m_1}$$ is $$9:7$$, which corresponds to Option C.