When applying the equation of torque and equating it to $I\alpha$ which moment of inertia do we take? I believe $T=I_{cm}\alpha$, where $I_{cm}$ is the moment of inertia about centre of mass and $\alpha$ is the angular acceleration.
But do we take $I_{cm}$ even if the torque has been taken about a point which is not the centre of mass? Would we then apply parallel axis theorem and find the new moment of inertia and use it in our equation? If no, then why not? 
 A: When it comes to dynamics always use the center of mass as as reference point.
$$ T_{\rm CM} = I_{\rm CM} \alpha \tag{1} $$
is what you want.
Well actually in vector form
$$ \vec{T}_{\rm CM} = I_{\rm CM} \vec{\alpha} + \vec{\omega} \times I_{\rm CM} \vec{\omega} \tag{1} $$
To transfer the above to some other point A, note that $\vec{F} = m\,\vec{a}_{\rm CM}$ and thus
$$ \vec{T}_A = \vec{T}_{\rm CM} + \vec{c} \times \vec{F} $$
where $c$ is the position of the center of mass relative to A, and
$$ \vec{a}_{\rm CM} = \vec{a}_A + \vec{\alpha} \times \vec{c} + \vec{\omega} \times (\vec{\omega} \times \vec{c})$$
Alltogether you have the equations of motion at an arbitrary point as
$$ \vec{F} = m  \, ( \vec{a}_A + \vec{\alpha} \times \vec{c} + \vec{\omega} \times (\vec{\omega} \times \vec{c}) ) $$
$$ \vec{T}_A = I_{\rm CM} \vec{\alpha} + \vec{\omega} \times I_{\rm CM} \vec{\omega} + \vec{c} \times \vec{F} $$
As you can see, not using the center of mass makes this much more complicated.
A: The moment of inertia is definitely affected by where the axis of rotation is located.  To find the torque required to rotate an object where the axis of rotation is not through the center of mass, you definitely need to use the parallel axis theorem.  If you intend to rotate a real-world object in such a fashion, expect a lot of "wobble" in the object if you intend to rotate it at any substantial angular velocity.
