Inductor's back-emf vs voltage drop across resistor

First of all, don't mix up voltage with current. In your examples 1 and 2 it is certainly true that the voltages across the resistor and inductor are the same w.r.t. the source voltage. This is just Kirchhoff's voltage law. However, this still results in a current lag in the inductor compared to the resistor. Say the source voltage is

$Delta V_S = V_0 sin(omega t)$

So the resistor voltage is

$Delta V_R = - V_0 sin(omega t)$

so that $sum Delta V = 0$ as Kirchhoff's law requires. The same goes for the voltage across the inductor, $Delta V_L$.

But for the resistor we have Ohm's Law

$Delta V_R = IR$ so the current through the resistor is just

$I_R = -frac{V_0}{R} sin(omega t)$

But for the inductor we have

$Delta V_L = - L frac{dI}{dt}$.

So to get the current, $I_L$, you need to integrate $Delta V_L$ w.r.t. time so in this example you will get a cosine instead of a sine. Thus, we see a phase shift in the current (but not the voltage). It is worth going through the integral yourself, but it is also in most elementary circuits textbooks.

This question and your explanation are total nonsense...

First of all, an inductor generates reactance not resistance, which is why in the frequency domain, we have an "i" in the equation for an inductor's impedance. This is where the 90 degree phase shift comes from. This is a basic manipulation involving complex numbers. This comes from taking a Fourier transform of Faraday's law with self-inductance.

Second, the back-EMF generated across the inductor depends on the frequency of the AC signal, which is why frequency appears in the equation for the reactance of an inductor. So you can't compare what happens in a resistor (which is frequency-independent) to what happens in an inductor (which is frequency-dependent).