1/2) More or less, I try to clarify, It seems to me you believe your graph in figure (2) shows R and L variations w.r.t frequency.
That graph simply shows impedance of a parallel resonant circuit made of approx constant 100nH inductor, 45kohm resistor and a \$\frac{1}{(2\pi\times 2\,\text{GHz})^2\times 100\,\text{nH}}\approx 65\,\text{fF}\$ capacitor.
This is one first level approximation which may however be good in many circumstances.
Then one can argue resistance is frequency dependant and add this to the model, the same for inductance and capacitance but this can usually hardly be seen on graphs unless you carefully measure, analize and fit curves to models.They are ususally hidden in measurements errors.
Then you can add many other extra parasitics and non linerities outlined above by other contributors, but there's no such an evidence in the sweep you posted.
b) Your meter shows variable (possibly negative) inductance just becouse you ask it to do so. It just measures an impedance, one complex ratio between voltage and current and then represent it as you configured it to do: you asked for Ls+Rs and if measured impedance phase does not agree with that model it just keep computing and findes "negative inductances".
Usabilty boundary (green/blue transition in fig. (2)) is not function of your component parasitic only, but it depends pretty much on the rest of the circuit e.g. if using that inductor in a resonator you should add stray capacitance to the computations and see if you get consisten numbers.
4) Yes SRF is that blue to red boundary.
3) Dimension do count. For resistors and capacitors usually the longer the part the higher the stray inductance. E.g. small SMD 0204 or 0603 may exhibit a few hundred pH while some big HV MKM capacitors I happened to use were specified as 7nH/cm w.r.t to terminals pitch.
In that scenario, a current would never flow.
Of course current flows through a capacitor each time the pulse rises and falls. For a capacitor Q=CV and differentiating on both sides to obtain current we find that: -
I = C dv/dt because dq/dt = current.
In other words when the voltage edge is sharp (high dV/dt) the current can be very large if not limited by some series resistance or an inductor.
It is for this reason that the current into a capacitor from a sine wave voltage source is shifted by 90 degrees into a cosine wave - it's all down to the maths behind differentiation of voltage (dv/dt).
Best Answer
Shunt stray capacitances are problematic when their impedance is too low, as opposite to being too high, because they pull down signal lines and/or mismatches them.
When the stray capacitance value is much lower than the frequency, then the resulting impedance may be high enough so as not to heavily impair a line. Everything might be reasonably under control.
However, as frequency goes up this may cease to be true, and then the resulting impedance may start to be low enough so as to pull down the line.