1. Does the electret 2-wire microphone require connection to a power source, to the 9V battery? I've heard both Yes and No.
The electret mic has a Jfet transistor as an impedance converter inside. It must be powered. Usually it is fed about 0.5mA (a current, not a voltage) through a 10k resistor from a filtered voltage of about 9V.
You have it powered through a 2.2k resistor that might reduce its level. The 2.2k resistor is not fed from a filtered volage so the amplifier will probably oscillate since the supply voltage will drop when the amplifier works hard which is coupled directly to the input of the amplifier though the 2.2k resistor as positive feedback.
2. I was planning to wire the + side of the microphone to both signal and +5V on a stereo jack. Will that work?
What is "signal"? If it is the output then you are shorting out its output signal to the positive supply.
It needs a resistor to power it so that the resistor is its load. A dead short is never a load.
3. The 0.0082 uF capacitor and the 10K resistor, next to the microphone, were my attempt to get a high-pass cut-off frequency of 1940 Hz. Did I succeed?
Yes it is a simple highpass filter. Simply calculate its cutoff frequency.
4. The datasheet claims that the IC will give 2W at 9V, when R(L) = 16 ohm. Is R(L) the resistor nearest the headphones?
No. The output jack with a speaker plugged in is the load.
An opamp can barely drive 1k ohms. The datasheet calls the Japanese IC a power amp not an opamp. Will you swim to Japan to buy one?
5. I'd like to use the mass-produced 8-ohm stereo headphones. Will 2.7V and 2W drive these?
Its output is 2 Whats at a horrible 10% distortion. Its output with a 9V supply is about 0.77 Watts into
an 8 ohm speaker at clipping. 0.77 Watts into headphones will probably destroy them and they will be so loud that they will destroy your hearing. Look in Google for Are Headphones Too Loud?
You have a circuit for opamps, not for this power amp. Most of your feedback resistors are already built-in.
Again, I want to use stereo headphones for a monophonic application. Will wiring the output to both headphones require any changes in the circuit?
You are using the "BTL" circuit with an amplifier driving each end
of a speaker for a high ouput power. Then the output power is much too high for headphones. Use a simple amplifier that drives only one end of headphones.
Maybe you should use an LM386 single power amplifier IC to drive your headphones at a reasonably low power but still pretty loud.
6. I feel that the gain is somehow controlled by the 4 resistors, 130R and 5K, fed from the IC's output. I'd like to introduce a potentiometer, as a kind of volume control. Where would I put it?
You have a circuit for opamps but you do not have opamps. A volume control is always a voltage divider at the input of an amplifier.
No.
You have the wrong amp.