Espruino Wifi battery?

Q1: Was this distinction VBUS vs VUSB made for schematic software reasons and not a typo?

There's no specific reason for that - it's because the design started off as the Pico design and had power supply components removed.

Q2: What/Where is the connector showing VUSB/DM/DP or is this the input to the WiFi module?

They're test points on the rear of the board

Q3: Under the assumption the micro USB connector is feeding the power, is it true that VBUS [ J2-2 ] is available as an output and that it is electrically connected to the input (after Fuse F1 perhaps) of the regulator as 5v (the source a USB port supplies), and therefore also supplying 3.3v as VDD [ J2-3 ] as an output?

VUSB is connected directly to the USB power supply, with no fuse. This was probably a bit of a mistake, but was done because people had complained in the past that there was no way to get the full power output from a USB wall supply/phone charger when using something like the Pico.

The regulator supplies 3.3v, but there won't be any problem supplying 3.3v from elsewhere - unless for example you're using a battery that is less than 3.3v (in which case it will be charged up to 3.3v when USB is connected).

Q4: Under the assumption that supplemental power from a 3.3v battery is supplied, and that no micro USB connector is attached, that is could be applied as an input VDD at [ J2-3 ] bypassing the onboard regulator?

yes. as I'd said in the last post - supplying regulated 3.3v to the 3.3v pin would be the most sensible way to power the device.

Note: This would mean there is no fuse protection and that input voltage would not be detectable at PA9/P30

Yes. If no input volatge is detected, the USB hardware won't be powered up, so the current draw will be less.

Q5: Similarly as above, upto 5v could be supplied by a regulated supply to VBUS [ J2-2 ], thus using the onboard voltage regulator, producing the 3.3v to the rest of the circuitry, and at VDD [ J2-3 ] as an output?

Yes, you could do this - again as I said in my last post. However there are some issues:

• The USB peripheral on the chip will stay powered, so will draw more power in sleep mode
  
• If you do that you cannot use the USB, since plugging it in will start to supply power into whatever power source you used (you'd need to connect the power source with a diode, and ensure it's always less than 5v)