Could there be a pseudovector kinetic term for fermions? Could there be a kinetic term of the form $\bar{\Psi} \gamma_5 \gamma^\mu \partial_\mu \Psi $ in addition to the usual one? Or is this forbidden by a symmetry?
 A: Theoretically, you are permitted to have such a pseudovector kinetic term : it's not forbidden by any symmetry.
Let's say the original vector kinetic term is
$$
\bar{\Psi} i \gamma^\mu \partial_\mu \Psi.
$$
And we add a pseudovector skinetic term
$$
\epsilon\bar{\Psi} i\gamma_5 \gamma^\mu \partial_\mu \Psi
$$
to get
$$
\bar{\Psi} i \gamma^\mu \partial_\mu \Psi + \epsilon\bar{\Psi} i\gamma_5 \gamma^\mu \partial_\mu \Psi
= (1+\epsilon)\bar{\Psi}_L i \gamma^\mu \partial_\mu \Psi_L + (1-\epsilon)\bar{\Psi}_R i \gamma^\mu \partial_\mu \Psi_R.
$$
The catch is that it's not how the physical world is ordained to be by the Almighty. An additional pseudovector skinetic term to the vector one would mean different kinetic terms (hence different momentums) for left and right handed fermion, which does not conform with experimental observations so far. 
At this point, the smart and sassy student in the front row might ask: "Hey professor, can't you just rescale the fermions fields in the Lagrangian to make the left/right-handed kinetic terms equal after all?"
Let's do the exercise of rescaling as
$$
\Psi_L \rightarrow \frac{1}{\sqrt{1+\epsilon}}\Psi_L.
$$
$$
\Psi_R \rightarrow \frac{1}{\sqrt{1-\epsilon}}\Psi_R,
$$
resulting in
$$
\bar{\Psi} i \gamma^\mu \partial_\mu \Psi + \epsilon\bar{\Psi} i\gamma_5 \gamma^\mu \partial_\mu \Psi \rightarrow \bar{\Psi} i \gamma^\mu \partial_\mu \Psi,
$$
which effectively kills off the pseudovector skinetic term and get us back to the original pure vector kinetic term.
How about the mass term? Dirac mass term would be rescaled as
$$
m\bar{\Psi}\Psi \rightarrow \frac{1}{\sqrt{1+\epsilon}\sqrt{1-\epsilon}}m\bar{\Psi}\Psi.
$$
Does this mean that there is no real physical effect of a pseudovector kinetic term, other than a rescaled mass term? 
The thing is that the gauge couplings will be effected. let's suppose that the fermion is coupled to a vector gauge field,
$$
\bar{\Psi} i \gamma^\mu (\partial_\mu -eiA_\mu) \Psi + \epsilon\bar{\Psi} i\gamma_5 \gamma^\mu \partial_\mu \Psi
= (1+\epsilon)\bar{\Psi}_L i \gamma^\mu (\partial_\mu -\frac{1}{1+\epsilon}eiA_\mu)  \Psi_L + (1-\epsilon)\bar{\Psi}_R i \gamma^\mu (\partial_\mu -\frac{1}{1-\epsilon}eiA_\mu)  \Psi_R.
$$
Let's apply the above mentioned rescaling of left/right-handed fermion fields and get
$$
\bar{\Psi} i \gamma^\mu (\partial_\mu -eiA_\mu) \Psi + \epsilon\bar{\Psi} i\gamma_5 \gamma^\mu \partial_\mu \Psi \rightarrow \bar{\Psi}_L i \gamma^\mu (\partial_\mu -\frac{1}{1+\epsilon}eiA_\mu)  \Psi_L + \bar{\Psi}_R i \gamma^\mu (\partial_\mu -\frac{1}{1-\epsilon}eiA_\mu)  \Psi_R.
$$
Oops, now we have both a vector gauge field
$$
\frac{1}{2}(\frac{1}{1+\epsilon} + \frac{1}{1-\epsilon})A_\mu
$$
and a pseudovertor (axial) gauge field
$$
\frac{1}{2}(\frac{1}{1+\epsilon} - \frac{1}{1-\epsilon})A_\mu
$$
Pseudovector gauge interaction is a can of worm you don't want to open. Apart from lack of experimental evidences, pseudovector gauge interactions will run into complications with quantum chiral anomaly cancellation considerations.  

A bonus for ya:
You CAN have both scalar and pseudscalor mass terms though, parameterized as:
$$
m\bar{\Psi} e^{\theta i\gamma_5} \Psi = m\cos\theta \bar{\Psi} \Psi + m\sin\theta \bar{\Psi} i\gamma_5\Psi.
$$
The fun fact is that after a rotation of the fermion field
$$
\Psi \rightarrow e^{-\frac{1}{2}\theta i\gamma_5} \Psi.
$$
the "complex" mass term can be rotated into a scalar mass term:
$$
m\bar{\Psi} e^{\theta i\gamma_5} \Psi \rightarrow m\bar{\Psi} \Psi.
$$
Interestingly, contrary to the earlier case of left/right fermion field rescaling, this rotation would not change the gauge couplings. 
