Why is only 2.8V from a 12V battery considered to be good??
The backwards base-emitter is a zener diode.
I looked up the pinout of the 2SC8550:

 

Why is only 2.8V from a 12V battery considered to be good??

You have only a small part of this circuit. I am working on a full schematic.
The battery is actually a 7S lithium ion. The 12V comes from a 78L12 on this board.
There is also an LM324 and the LED circuit input comes the output of one of the op-amps in that chip.

 

Attached is my first draft of the schematic for this board. It is weird.
None of the opamps (LM324) have feedback so they are being used as comparators. One of them has no connections at all.
Two of the outputs are connected by a diode so they will be in conflict in certain situations.
What is R12 for?

The shunt resistor 0.15Ω is open circuit, R14 is burnt and open circuit, The 78L12 is short input to ground, the LM324 is faulty.
I doubt anything else is damaged so it would be easy to repair.

Because it is so weird I am debating replacing this board with an 8 pin PIC version which would be easy to do and easy to fit onto the same size board - considerably fewer components.

[Edit] IN is the charger - 7S lithium charger, OUT is the battery, the diode is a 1N4148

 

Here is the section of the board showing these transistors.
View attachment 267270

Which does not help since the orientation is not indicated.

Bob

 

The shunt resistor 0.15Ω is open circuit, R14 is burnt and open circuit, The 78L12 is short input to ground, the LM324 is faulty.
I doubt anything else is damaged so it would be easy to repair.

I have now replaced these parts and it is working again - hurrah!