I won't solve the question for you, but I guess a clue will help you better to grab the concepts rather than some complicated solutions.
Here you see we will have to apply the external force F if friction isn't enough to keep it from moving. So let the friction do the most it can, i.e. apply the maximum value of friction (coefficient of friction times normal reaction) and the rest is done by an external force F.
The force F results in motion of the whole system and the acceleration would be $\frac{force}{totalmass}$ which results to some pseudo force on the smaller body, if you are sitting on the larger block (to be complicated in the same frame as of the larger block) some component of this force adds to the frictional force and keeps the smaller block from motion.
Figure out the rest yourself.
First thing is, Friction prevents relative motion. What this means is, that the value and the direction of friction will so adjust, that it will try to minimize relative motion between the surface and the block and to prevent it, if possible.
Next, static friction is an self-adjusting force. What this means is that, its magnitude and value is not fixed but it will change depending on the different physical situations, i.e. different forces acting on the body. It's maximum value in this case will be $\mu_s N$ but it can have any value from zero to this maximum. The value depends upon the situation and is always such that it avoids relative motion. (having the maximum value is the best try friction can do to prevent relative motion, and beyond that, if the external force is increased, relative motion occurs and the value of static friction is substituted by kinetic friction.)
In your problem, you should first try and check if a stationary (zero acceleration) solution is possible, i.e. will friction be able to prevent relative motion (between the inclined plane and the block 2). To do this, draw a free body diagram of block $M_2$. It has $T$ tension acting upwards (along the incline of the plane) and $M_2g\sin\theta$ acting downwards (along the incline). Let us suppose a static solution is possible, i.e. friction is able to prevent all motions.
Then, both the blocks must be stationary. This means that the tension $T=M_1g$ since, on block 1 there is only the downward gravitational force which needs to be balanced by the upward tension if it has to stay stationary.
Going back to block 2, the forces along the incline are $M_1g=10N$ upwards and $M_2g\sin\theta=5N$ downwards and the friction. Since the upwards (along the incline) force is greater than that downwards, the friction must act in a downward direction to keep the block in equilibrium. The magnitude of friction needs to be $5N$ which is less than the maximum value of friction possible $\mu_sN=0.6M_2g\cos30=5.196N$, and hence the static solution is possible. Here, the frictional force will be $5N$ and not $5.196N$ as static friction self-adjusts to the value required to maintain a stationary state, i.e. prevent relative motion.
The 2 negative accelerations you get by assuming maximum static friction and then solving, indicates that a static solution is possible, since both accelerations turning out to be negative with maximum static friction means that maximum static friction is more than the other forces and hence friction is capable of balancing other forces and create a static solution.
Best Answer
If by "theta max" you mean the maximum incline angle where the block will remain at rest (not start to slide), then the answer is no.
The maximum incline angle before sliding is impending is independent of the mass of the object or the force of gravity. This can be shown as follows:
The component of the force of gravity acting down the plane is given by
$$F_{g}=mg\sin\theta$$
The normal force on the plane is $mg\cos\theta$. Thus the maximum possible static friction force acting up the plane is given by
$$F_{s-max}=\mu_{s}mg\cos\theta$$
Where $\mu_s$ is the coefficient of static friction between the block and the incline surface.
Impending sliding of the block is when the two forces are equal
$$mg\sin\theta=\mu_{s}mg\cos\theta$$
$$\mu_{s}=\tan\theta_{max}$$
$$\theta_{max}=\tan^{-1}\mu_{s}$$
So $\theta_{max}$ is independent of the mass and the magnitude of the force of gravity. It depends only on the coefficient of static friction. So the force of gravity, be it $g$ or $\frac{1}{6}g$ (approximate on the moon) is not a factor.
Of course if there were no gravity, the block will not move regardless of the inclination since there is no downward force along the incline nor normal force perpendicular to the incline (i.e., the net force acting on the block is zero).
Hope this helps.