Electronic – Op-amp voltage follower not working as expected

1. No, no reason for them to be equal, so long as you observe the maximum (and minimum recommended) total voltage and keep inputs within the common-mode range.

2. The minimum common mode range is +/-11V with a +/-15V supply. With supplies +15 and -8 you will have a common mode range of +11/-4. The input will work in this range, in the follower configuration.

3. Certainly not. The current is guaranteed to be less than 60mA. It's a minimum of only 10mA and that's with a 15V supply. Use the output swing numbers. With a +/-15V supply it can supply +/-10V to a 2K load (5mA). It might be a bit less with a 22V supply compared to 30V, but probably in that range.

4. Voltage followers are prone to oscillation if the load is capacitive, otherwise this amplifier is compensated to be unity-gain (the way you are using it) stable.

Your third image is not a (source) follower, rather it is a common-source amplifier and generally is configured to provide gain > 1.

A source follower looks like this:

Note that there is no drain resistor at the top and that the output is taken from the source terminal rather than the drain.

\$V_{out}\$ will follow \$V_{in}\$ with a drop of roughly the gate threshold voltage of the MOSFET, \$V_{th}\$. This may be somewhere around 5V depending on the device, so one might reasonable choose a BJT emitter follower instead:

You can see the circuit is essentially identical excepting the transistor type. \$V_{out}\$ will follow \$V_{in}\$ with a drop of \$V_{BE}\$, typically only 0.7 V.

The input impedance of an emitter follower is relatively high, and it's output impedance is relatively low. So placed between a resistive voltage divider and the rest of the circuit (such as an amplifier input) has the effect of stabilizing the bias voltage developed across the divider against variation due to changes in current drawn from the divider.