Acceleration of the centre of mass of a hinged rod 
This is a freely falling rod of mass m and length L hinged at one end. 
I have to find the acceleration of its centre of mass. 
1) I don’t think its acceleration can be greater than g in any case as there isn’t force available greater than mg. Is this reasoning correct? 
2) I don’t understand why its acceleration won’t be the vector sum of its rotational and translational accelerations here.
mg will provide a translational acceleration of g and an angular acceleration of 3g/2L. The angular acceleration will give the centre of mass a tangential acceleration of 3g/4. 
Why isn’t the net acceleration 3g/4+g? (Its acceleration being greater than g doesn’t make sense either though.) 
3) Shouldn’t the hinge force have something to do with this acceleration? If it does, maybe the acceleration can be greater than g then as the hinge force could add to it in some way? 
4) What will be the direction of the hinge force and why? I think it’ll be upwards at this instant and as it falls as all the hinge is doing is preventing the point where the rod’s hinged from falling down. 
 A: Based on your question and comments, it seems like you are confusing yourself with how forces relate to accelerations.
Forces produce accelerations. These accelerations have components. However, in circular motion when we differentiate between tangential and angular accelerations we are really just talking about the tangential component of the acceleration in two different ways. In other words, you don't split the acceleration caused by a force into a tangential part and an angular part. You just explain it in one way or the other way. The tangential acceleration looks at the change in tangential velocity over time, which relates to the distance traveled. The angular acceleration looks at the change in angular velocity, which relates to the angle swept out by the radial vector.
So, in your example, you have two forces acting on your rod. You have gravity and the hinge force. These two forces will produce an acceleration of the rod. How can we determine what this acceleration is? Well, let's be smart about it. We can look at the torque caused by gravity about the position of the hinge. This is useful because how to directly handle the hinge force is tricky, but we don't need to worry because the torque of a force about the point that force is applied is $0$. Also, we know how gravity behaves, and it is a simple matter to determine the torque of gravity about the hinge.
So, let's use Newton's second law:
$$\tau=I\alpha$$
where $\tau$ is the net torque about the hinge acting on the rod, $I$ is the moment of inertia of the rod about the hinge, and $\alpha$ is the angular acceleration of the rod about the hinge (notice how I am not saying "the torque caused by so and so force" or "the angular acceleration caused by the so and so torque". We have a net torque, we have an angular acceleration. That is it.
Now, since we already determined $\tau_h=0$, we have for a uniform rod, $$\tau=\tau_h+\tau_g=0+\tau_g=\tau_g=\frac12mgL$$
and the moment of inertia of the rod about its end is $I=1/3mL^2$ This gives us what you were expecting:
$$\alpha=\frac\tau I=\frac32\frac gL$$
Note that this is the angular acceleration for all points on the rod. However, if you want to know the tangential acceleration (same acceleration, just displayed differently) of a point on the rod a distance $R$ from the hinge, you have $a=\alpha R$. For a point at the middle of the rod, $R=1/2 L$, therefore
$$a=\frac34g$$
which is what you wanted to know.
Please note that this is not a solution to a homework problem, since you already knew the answer. I have supplemented steps with understanding that you can hopefully apply to other systems you are looking at. I will say it again, trying to figure out what to do with the hinge force is tricky. At the same time, you don't break the acceleration due to gravity into a tangential and angular part. You also don't break gravity into a force and a torque. Forces have torques about any point you choose. Angular accelerations are just tangential accelerations expressed in terms of angle covered instead of distance traveled. 

In summary, to explicitly answer your questions,

1) I don’t think its acceleration can be greater than g in any case as there isn’t force available greater than mg. Is this reasoning correct?

I am unsure about this. I wouldn't worry about it here though. Just determine the accelerations.

2) I don’t understand why its acceleration won’t be the vector sum of its rotational and translational accelerations here.

This is because the rotational and translational accelerations are not the two orthogonal components of the acceleration vector. The translational (or linear, tangential etc.) acceleration is one component. The radial acceleration is the other component. But the angular (or rotational) acceleration is not a vector component of the total acceleration. It is just the tangential acceleration expressed per unit length from the center of rotation.

3) Shouldn’t the hinge force have something to do with this acceleration? If it does, maybe the acceleration can be greater than g then as the hinge force could add to it in some way?

Absolutely. The hinge force does influence the acceleration of parts of the rod. However, we were smart above by using the fact that the hinge force has no torque about the location of the hinge. So we didn't need to care about what the hinge force was actually doing here (although you can determine it knowing how the center of mass moves as the bar swings. I will leave this to you).

4) What will be the direction of the hinge force and why? I think it’ll be upwards at this instant and as it falls as all the hinge is doing is preventing the point where the rod’s hinged from falling down.

The hinge force will have a component pointing along the rod to allow for the rod to move in a circle, but it will also have a vertical component to support the left end of the rod.
A: The net force acting on your rigid body , the rod, is $\vec F_{\rm gravity}+\vec F _{\rm pivot}$ with the force due to the pivot, $N$, in an upward direction and that due to the gravitational attraction of the Earth, $mg$, in a downward direction.  
The linear acceleration of the centre of mass of the rod, $a$, will be given by $mg-N = ma \Rightarrow a =g-\frac {N}{m}$ ie less than $g$.  
The $g$ which keep wanting to include is the acceleration of free fall, when $F_{\rm gravity} = mg = ma \Rightarrow a=g$ and in your example the rod is not in free fall but constrained by a force at the pivot.
