Electronic – Power Consumption and efficiency

I can help to address the first part of your question. As an abstraction of a single switching device inside a processor, imagine a MOSFET connected to ground with a load resistor to the power supply (in a real CMOS part there wouldn't be a load resistor, but another transistor, but this distinction is not important for the analysis). Connected from the junction of the resistor and the transistor is a capacitor, representing all the input capacitances of the transistor that the transistor under discussion is driving. When the first transistor switches off, this capacitance will be charged up through the load resistor. When the first transistor switches back on, the charge stored on the capacitor will be discharged through the first transistor.

It can be shown that when a capacitor is charged through a load resistor, 1/2 of the energy used in charging the capacitor is lost in the resistor, for a total energy dissipation of \$\frac{1}{2}C{V_s}^2\$, where \$V_s\$ is the supply voltage. When the switch then turns on, assuming the resistance of the switch is much less than the load resistance, that same energy will be dissipated in the switch, for a total energy of \$C{V_s}^2\$. Dividing this by the switching period gives you the dynamic power dissipation of the switch/capacitance combo, \$C{V_s}^2f\$. Shrinking the die reduces the junction capacitances of the MOSFETS, so if you know the supply voltage, switching frequency, number of transistors and the approximate junction capacitances of a certain process, you could calculate a ballpark figure of what kind of power savings a process shrink entails, all other things being equal.

Q1: There is some internal circuitry within opamps (bias, ...) whose current consumption depend on the provided voltage. So, even if it does not change anything to the opamp output, changing its supply voltage can change the current it internally draws (called the quiescent current).

Q2: You can't, with this model (see below)

Q3: It is indeed inconsistent. If the opamp sources 9mA to its output, the positive supply current should be at least 9mA (plus the internal quiescent current). And if it sinks XmA, the negative supply current should be -XmA. In real life, that's how it is. Now, in simulation models, most often, this is not taken into account. Most manufacturers publish opamp models that are inaccurate regarding supply current, because it simplifies their models a lot, and supply current is not usually checked during simulations.

In short: to check the opamp consumption, you, most often, have to do it manually: compute the power used to sink/source the required output current, and add the power used from the quiescent current given in the datasheet (which is always provided).