How does this Op-Amp Circuit Work?
All three of your questions are related.


This serves a dual (triple? see #2) purpose of blocking any DC voltage on the line, and re-referencing the voltage to something the circuit can use.
Note that the voltage on the + pin of U1G1 is connected to Vcc/2. If you apply your laws of op amps, you know that in negative feedback mode (where the output is connected to the - pin through feedback) the + and the - port will be equal voltage. This MIGHT cause current to flow into the output of whatever you've plugged in to JP2, but the capacitor prevents that.
The output capacitor is an easy way to prevent DC voltages from appearing on downstream amplifiers (which would amplify the error voltage, causing bigger offset problems), as well as preventing current flow in speaker coils that are attached which at best burns energy and at worst burns up the speaker coil!


DC error/offset in the output is an inevitable effect of reality in most amplifiers. There are some designs (called DC servo designs) that have a feedback loop to cancel this output DC offset. This reduces the need for output capacitors in power amplifiers, but most of the authors I've read are pretty down on the idea as a practical solution to the offset problem.

A note on filtering: You're probably looking at this like an RC circuit, but at the frequencies we're looking at, the source impedance is low enough and the capacitance is high enough that the effect is minimal. Generally the larger the capacitor, the better the bass response and lower distortion.