Below is a diagram based off the OP. I hope it provides some more clarity, and is not just a repeat of information:

https://www.dropbox.com/s/x3lfg90f15ust1f/011016_Pulley System.gif?dl=0
Q: We want to solve for acceleration of the two masses

Answer:

Sign convention

The sign convention is:

Moving in the positive direction means (as shown by image in OP): up from the right hand side and around and then down toward the ground on the left hand side

System 1 (LHS)

Let a be the net acceleration of the pulley system 1.

System 2 represents everything on the right hand side of pulley 1, and system 1 represents everything on the left hand side of pulley 1.

The net force for system 1 is given by:

EQUATION 1: mass1*a = T1 – (Mass1)*g

System 2 (RHS)

We have T2 = (mass2)*g, since the chord for pulley 2 is fixed at one end. We have, following the same sign convention as pulley 1:

(Mass2) – T2 = 0

T2 = Mass2

T2 - Freaction = 0

And (Mass2) – Freaction = 0

(Mass2) = Freaction

This implies that the total weight for system 2 is:

W2 = T2 + T2 = 2*(mass2)*g

And the net force is (let a be the net acceleration of the pulley system 1):

EQUATION 2: (mass2)*a = W2 – T1 = 2*(mass2)*g – T1

NOTE: The acceleration of all objects within system 2 is 0. Imagine a bus with all things inside bolted down to cancel inertial forces (or inertial dampeners if you watch Stargate). The entire system 2 can be accelerating whilst its contents do no accelerate. So pulley 1 causes system 2 to accelerate (this is possible because pulley 2 itself should be able to move up and down).

Total system

Net force on pulley system 1 = (Mass1 + Mass2)*a = EQUATION 1 + EQUATION 2 = 2*(mass2)*g - (Mass1)*g; (the tensions cancel out)

We can now solve for acceleration:

a = [2*(mass2)*g - (Mass1)*g]/(Mass1 + Mass2)

[2*(5 kg)*(9.81 m/s^2) – (67*9.81 m/s^2)]/ 72 kg = -7.76625 m/s^2 (Do your rounding after all calculations complete to minimise error; use this value for calculations)

Therefore acceleration equal -7.77 (m/s^2). The sign convention means that the masses move in the same direction (up from the right hand side and around and then down toward the ground on the left hand side).

Extra notes

To find T1 substitute the acceleration into equation 1 and solve for T1. This should give 136.93 N.