Output resistance:
Roughly between 10 and 50 ohms, depending on part, unit, temperature and so on. It's also non-linear because the die locally heats up under load I^2*R, which in MOSFETs, as you may know, increases the R by the positive tempco. This happens very quickly.
It's OK for powering some very low-power loads. LEDs are the obvious typical culprits.
100nF bypass cap is, in my opinion/experience, OK, inrush currnet is theoretically high but the duration of it is short.
The result of this is that whatever you are powering from the IO pin must not try to pull significant peak currents.
Input resistance:
Nearly infinite, tens of megaohms, plus maybe some tens of pF of capacitance, not much more than a small piece of wire itself. No current flows into the input pin. HOWEVER, this only applies when the input voltage is within the voltage rails of the device (or slightly outside, like between GND-0.3V to Vcc+0.3V). Very important consideration when the device is still powered off. Otherwise, internal ESD diodes conduct, powering up the Vcc through that input. The protection diodes can take a few mA continuous, only. If the MCU itself is low-power, it may survive such abuse, i.e., being powered through IO pin protection diodes. Otherwise than that, you want to limit that protection diode current to, say, below 1mA. Then just do the Ohm's law to obtain the series resistor value you need. Do note that even just 1mA may cause the MCU to power up, boot and do Funny Things, so you may need to limit the current further.