Is it easier to push or pull a 100 kg mass up a 30 degree angle

The coefficient of friction is 0.4

In both cases I am pushing/pulling with a force of 600N

1 - Find the net force of pushing the mass
2 - Find the net force of pulling the mass

In either case the friction will act to oppose the motion, so my best guess is that the two answers will be the same.

The frictional force in each case is mu F sin theta = mu mg sin theta, the frictional force times the component of the weight force of the mass that is acting 'normal' (at right angles to) the ramp.

Therefore the frictional force is F(fric) = 0.4 * 100 * 9.8 * sin 30 = 196 N

Now, here's where the question gets tricky: which way is the 600 N force acting? Is there a diagram?

If both the pushing and pulling force are parallel to the ramp, then the net force is simply 600 - 196 = 404 N in each case.

If the pulling force is vertical and the pushing force horizontal, for example, you will need to first calculate the component of the force that is acting parallel to the ramp, and then subtract 196 N from that. It is likely to be either 600 cos theta or 600 sin theta: you need a diagram.

That's as far as I can take it without more information but it should be enough to get you going.

There is no difference.
The net force is (600 - 100*cos 30*0.4) in both cases.