Inconsistencies between the Nortons and Thevenins equivalent circuits

A Thevenin or Norton equivalent is a circuit that has the same I-V curve as some more complex linear network, but itself contains only two elements: a source and a resistor or conductor.

In your first circuit, resistor R1 is irrelevant to finding the Thevening equivalent. Since R1 is in parallel with a voltage source, the overall I-V curve is just a vertical line through 1 V, as if you had only a voltage source. Which is exactly what your Thevenin equivalent circuit shows.

The reason you can't find a Norton equivalent circuit is because a Norton circuit can't produce a vertical line I-V curve with finite element values. You could create (or at least think about) a Norton equivalent circuit in limit as \$I\to\infty\$, if you maintain \$G=\frac{I}{1\mathrm{V}}\$ as you increase I.

The second circuit is just the dual. It has a horizontal I-V curve, and a Thevenin circuit can't have a horizontal curve with finite values.

The problem is not to have one (or two) variable resistors, because you always take a look at one moment and treat them as fixed during your analysis.

I guess your problem is to see that the variable resistor (potentiometer) actually consists of two resistors (R2 and R - R2). In general they form a voltage divider; in this particular case one of the two "sub"-resistors (R - R2) is shorted so you can forget the (R - R2) part (i.e. treat it as being 0).