The microphone's output impedance is irrelevant to the choice of op-amp, because you "program" that aspect by a suitable op-amp circuit.
The low impedance of the mic means that the amplifier can have a low input impedance, in the thousands of ohms. But if the connection from the mic to the amplifier is short (we don't have to worry about stray capacitance of a cable), it doesn't have to. You can build the amplifier to have a relatively high input impedance, like 50 kOhms and up.
If you plan on using a coupling capacitor, like in the recommended circuit, a low input impedance will work against you: a low R means you will need a large C to maintain frequency response, which is linked to the RC product. (Since you give the audible range as 10 Hz (!) to 20 kHz, it can be assumed that you care about low frequency response).
The choice of op-amp depends on various parameters. This is a shopping question that is generally considered off-topic (on most stackexchange sites). You probably want it to be a low-noise unit suitable for audio, which has published distortion figures which are low. Then you have to consider your power supply: would a dual op-amp IC that drains up to 16 mA of current be acceptable? Or how about one that needs a minimum of 10V across its power rails to work properly: would that work? Cost: is it okay if the op-amp costs ten dollars? Or is fifty cents more appropriate? Output: does the op-amp have to produce output that goes almost all the way to the power rails? Or is it okay if it only goes to within a few volts of either rail before clipping? Manufacturing: are you comfortable with small, surface-mounted IC's, or would it be better to have a classic through-hole part with 0.1" pin spacing?
Whether or not a voltage divider is the best approach to power the mic depends on how much wattage will be wasted, and whether you can afford it.
Use a rail to rail opamp that can run from 3.3 V. Many of the Micropchip MCPxxx line, for example, can do what you want.
Run the opamp in positive gain configuration with ground as the reference. The gain needs to be about 8. All you should need is the opamp with its bypass cap, and two resistors.
As Andy pointed out, even "rail to rail" opamps might not do what you want within a few 10s of mV of either rail. Check the datasheet.
If this is a problem, you can use a small charge pump to make a small negative voltage. There are charge pump ICs for this purpose, but if you have a microcontroller with a spare pin this can be even simpler. I have used the clock output feature of a micro a few times just to run a charge pump from. If you only need a few mA, you can run the charge pump directly from the clock out or PWM logic signal. All you need is two diodes and two caps. If you need a little more current, follow the logic signal with a NPN/PNP emitter follower pair. That will reduce the negative voltage magnitude by 1.4 V or so, but even with 3.3 V supply there is plenty left to get ground well past the rail region of a "rail to rail" opamp.
With a volt or two negative supply, you can sometimes use much cheaper opamps, which makes the charge pump option cheaper overall. For example, if you can use a LM324 if only you had a negative voltage, then quite likely the more expensive opamps that go to the negative rail will outweigh the few diodes and caps to make the charge pump.
As always, look at the opamp datasheet carefully. If the micro is running from 5 V and you are using the logic signal directly to drive the charge pump, you may actually be giving the opamp too much voltage. The opamps that get close to real rail to rail performance also tend to work over a limited supply range.
If you do use a charge pump, it would be a good idea to filter the charge pump output before it is applied to the opamp negative power input. A ferrite "chip inductor" in series followed by a 20 µF cap to ground should be enough.
Best Answer
It doesn't need a dual supply. It just needs to operate in a space that has some distance between the negative and positive rails, and for most common power supply schemes, that space is pretty small if the negative rail is at ground, and the arrangement isn't very useful.
As to "why" -- it's the configuration of the transistors in the device. I'm not sure you're interested in the gory details, and I haven't checked them, but often this has a lot to do with bias voltages used to keep transistor amplifiers in range for linear operation. When the chip was designed, just having access to a device as facile as an op amp was a pretty good thing. These days, we want less batteries with more life, less power, smaller voltages, and an op amp designed so long ago just doesn't fill the niche anymore.
If you look at the datasheet at http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=UA741&fileType=pdf, you'll see in section 7.3, the \$V_{ICR}\$ line, that the common mode input needs to be about 3 volts above the negative rail and 3 volts below the positive rail. The \$V_{OM}\$ line shows a similar range for the output. Note that this means the chip is entirely unsuitable for power rails at 0V and 5V, but there is a bit of room if you're operating around 5V for rails at ground and 10V.
So, the "why" isn't there, but the "will my op amp work the way I want it to" is right there for anyone who knows how to find and read the instructions.