Does the frequency of AC current affect the induced emf in a transformer (mutual induction)? Faraday’s law states that the induced emf is directly proportional to the rate of change of flux.
So if I have a constant input voltage and increase the frequency of an alternating current in a transformer, would the rate of the secondary coil cutting field lines and thus the emf induced increase and if not, why?
I’m a bit confused by the word “rate”, and the idea of frequency affecting voltage seems really absurd to me.
 A: 
would the emf induced increase and if not, why?

When you have two coupled (ideal) inductors, you have to consider a coupled set of equations rather than one.
For example, and assuming sinusoidal excitation at angular frequency $\omega$, we can write
$$V_p = j\omega(L_p I_p + M I_s)$$
$$V_s = j\omega(M I_p + L_s I_s)$$
where $L_p, L_s, M$ are the primary, secondary, and mutual inductances respectively.
Consider the case that the secondary is open (not connected to a load) so that $I_s = 0$.  Then
$$V_p = j\omega L_p I_p$$
$$V_s = j\omega M I_p$$
and it follows that
$$V_s = j\omega M\frac{V_p}{j\omega L_p} = \frac{M}{L_p}V_p$$
The angular frequency dependence has canceled out!  In the case of unity coupling, (no flux leakage), we have that
$$M = \sqrt{L_p L_s}$$
and so
$$V_s = \sqrt{\frac{L_s}{L_p}}V_p$$
Since the inductances are proportional to the square of the number of turns $N$, we finally have the ideal transformer relation
$$V_s = \frac{N_s}{N_p}V_p$$

When the secondary is connected to a resistive load, Ohm's law yields
$$V_s = -R I_s$$
(The negative sign is there because we take $I_s$ to be positive if it is in to the more positive terminal of the secondary) and the second of the coupled equations becomes
$$-R I_s = j\omega(M I_p + L_s I_s) \Rightarrow I_s = -\frac{j\omega M}{R + j\omega L_s}I_p$$
Note that in the limit that $R \rightarrow 0$ and unity coupling, we get the ideal transformer relation
$$I_s = -\frac{N_p}{N_s}I_p$$
But for non-zero $R$, the secondary current is a function of the angular frequency $\omega$.  Once again, assuming unity coupling, we get 
$$I_s =  -\frac{j\omega \frac{\sqrt{L_p L_s}}{R}}{1 + j\omega \frac{L_s}{R}}I_p$$
Note that as $\omega \rightarrow 0$, the secondary current goes to zero (for finite $L_p, L_s$) and so, there is no transformer action at DC.
On the other hand, for $\omega \gg \frac{R}{L_s}$, we recover the ideal transformer relation.

Having solved for $I_s$ in terms of $I_p$, we solve the first coupled equation for $V_p$ and get
$$V_p = \frac{j\omega L_p}{1+j\omega\frac{L_s}{R}}I_p$$
Again, note that as $\omega \rightarrow 0$, we have
$$V_p \rightarrow j\omega L_p I_p$$
and see that there is no transformer action at DC, i.e., the primary doesn't 'see' the resistance attached to the secondary.
Also, for $\omega \gg \frac{R}{L_s}$, we have
$$V_p \approx \left(\frac{N_p}{N_s}\right)^2R\,I_p$$
which is the ideal transformer relation for the transformation of impedances.

Finally, note that in the dual limit as $L_p, L_s \rightarrow \infty$, keeping the ratio $\frac{L_p}{L_s} = \frac{N_p}{N_s}$ constant, the ideal transformer equations become exact over all $\omega$, i.e., the ideal transformer 'works at DC'.
