Electronic – Emitter resistor and biasing voltage

It depends on the bandwidth you want the amplifier to achieve. A simple thumb rule is that a capacitor is seen as a "short-circuit" at higher frequencies and an "open-circuit" at DC.

Therefore, the gain of a common-emitter is about the load connected to the collector divided by the load connected to the emitter. So, you can change the AC gain of the amplifier using a capacitor at the emitter in parallel without changing the bias point of the transistor. A .1 uF in parallel with a 1k resistor will put a low cut frequency of about 1.6 kHz. Therefore, higher values should be used for lower frequencies.

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.