First of all, understand that an ideal capacitor has infinite impedance to DC signals, so the two waveforms you drew will generate identical current across a capacitor (while they would generate very different currents across a resistor).
1) A concrete example: given a sine wave that is 1V peak-to-peak, 0 DC bias would be it varying from -0.5 to +0.5; with 1V of DC bias, it would range from +0.5 to +1.5. An ideal capacitor has infinite impedance to DC, so either of those waves would look identical to the input of this circuit.
Notice that R1 and R2 create a voltage divider. C1 not only prevents any DC bias from entering the circuit, it also prevents this DC bias voltage from appearing to the signal source.
This bias voltage is needed in order for the transistor to work at all (for there to be collector-emitter current, there must be a smaller current from the base to the emitter). What will happen is that the base-emitter current will be proportional to the AC component of the input signal, plus the DC current established by the path from R1, through the base-emitter junction, to Re.
1a) (describing how Ce works) Remember that this amplifier works by having the base-emitter current (small) control the collector-emitter current (large). Without Ce, both of those currents would always have to pass through Re, which limits the gain of the amplifier. Having Ce there means that any high-frequency components of the signal do not pass through Re (since a capacitor has zero impedance to a high-frequency signal).
1b) You didn't ask this, but Re is there in the first place to stabilize the biasing of the transistor. If no Re was needed, then Ce wouldn't be needed either (because the whole point of Ce is to allow AC currents to bypass Re).
Lastly, you might want to read the wikipedia page on BJT biasing. It's not perfect, but it walks you through all of the circuits that are simpler than the one you posted, until it finally gets to the one you posted, and it provides some (minimal) explanation for each step.
2) I'm not sure why they show that as a bypass capacitor. Bypass capacitors are used because real world circuits have resistance and inductance, so the large spikes in currents caused (in particular by digital signals, and especially with MOSFET digital logic) will cause the voltage to "sag" which can mess things up. Capacitors provide a very low impedance path for these currents, which keeps the voltage more consistent