Incline kinematics problem

mass 1 = 2 kg
mass 2 = 3 kg
Mass 1 is hanging by a weightless cord on a pully and mass 2 is on an incline with frictionless surface (theta = 60 degrees). Calculate acceleration of the system.

I got 5.7 m/s^2. Is that right? I used (1) FT or cord tension but I'm not sure if I need to use 2 FT in the equation for force.

You need a clear diagram. It should show the tension and weight acting on m1 and the tension and component of weight acting on m2.

The resultant force on m1 in its direction of motion (vertically upwards, taken as positive) is:
T - mg
= T - 2x9.81
= T - 19.62

The resultant force on m2 in its direction of motion (downwards along the incline, taken as positive) is:
mg.sinθ - T
= 3x9.81sin(60⁰) - T
= 25.49 - T
(mg sinθ is the component of weight along the incline - if you don't understand this, see near the end of the video-lesson in the link)

We can apply 'F=ma' to each mass separately to get 2 equations:
For m1, 'F=ma' gives: T - 19.62 = 2a
For m2, 'F=ma' gives: 25.49 - T = 3a

Adding the left and right hand sides of these equations gives
25.49-19.62 = 5a
a = 5.87/5
= 1.2m/s² (to 2 significant figures)