Electronic – Which of these Stripline Impedance Calculations is correct

I think you're on the right path, a couple of notes,

1) With a signal trace between two planes, the return current will split between the two planes, even if one of the planes is split. The return current cannot "see the future" and decide ahead of time which plane to return on. It will return above and below the trace until it sees the split at which point is says "oh crap!" and pays you back by possibly causing you to fail FCC testing. So you want to avoid running traces over plane splits even if another adjacent plane is not split. You can deal with splits with capacitors and such but this type of solution is less than ideal. I'd focus on always avoiding running a trace over a plane split on an adjacent plane.

2) Wide return paths on DC signals don't really matter.

3) You asked about two signal layers sharing the same plane. Usually, this is not a big deal if done properly. What many people do is use one of the layers as a "horizontal" signal layer and the other as a "vertical" signal layer so the return currents are orthogonal to each other. It is very common to route two signal layers for each plane, and use this horizontal/vertical technique. The most important thing to remember is to not change reference planes. Your setup could be a little tricky because going from the bottom layer to the 4th layer adds another return plane. More typical 6 layer boards are

1)ASignalHor 2)GND 3)ASignalVer 4)BSignalHor 5)POWER 6)BSignalVer

If you need smaller additional planes, like under the micro, these would usually be placed as an island on one of the signal layers. If you need to use more power planes, you might want to think about going to 10+ layers.

4) Plane spacing is important, and can have huge impact on performance, so you should specify this to the board house. If you take the example 6 layer stackup I mentioned above, spacing of .005 .005 .040 .005 .005 (instead of standard stackup with equal distance between layers) can make an order of magnitude improvement. It keeps the signal layers close to their reference plane (smaller loops).

I certainly hope I managed to decrypt what's being asked here. If not, please clarify so I can provide a more relevant answer.

Let's start with the definition of the lossy and lossless models:

In lossy model the resistance of the line itself matters for the calculation! Basically we're running such voltage through the line that even if we just took replaced the signal with a DC voltage equal to the RMS voltage of our signal, we'd still have losses.

Using the lossless model, if we replaced the signal with DC voltage equal to RMS voltage of the signal, we'd have no considerable losses. In this case, only reactive components matter and the resistance of the line itself isn't important.

This brings us to the basic equation for the transmission line characteristic impedance:

\$Z_0=\sqrt{ \frac {R + j \omega L} {G + j \omega C}}\$

Keep in mind that G is conductance in parallel with the capacitance, and that R is resistance in series with the inductance. In the lossless model, we don't have the two of them, so their values are taken as zero. We can then rewrite the equation as

\$Z_0=\sqrt{ \frac {0 + j \omega L} {0 + j \omega C}}\$

which is equivalent to

\$Z_0=\sqrt{ \frac {j \omega L} {j \omega C}}\$

Which we can rewrite as \$Z_0=\sqrt{ \frac {j \omega} {j \omega } \frac {L}{C}}\$

The \$j \omega\$ parts prodce one and we get in the end only

\$Z_0=\sqrt{\frac {L}{C}}\$

So the lossless model doesn't have the imaginary part because the imaginary units cancelled themselves out.