The questions has been solved!

 

The two transistors with double emitters are the inputs. Look at what they do when all four emitters are at a high level. And then when one or all gets pulled low. The transistors are set up as switches.

You have a fixed voltage supply and values for the resistors, so power is easy to calculate (don't forget the PN junction drops).

 

I got it, thank you very much, beenthere. I'm too tired to write the solution, so I'll skip it.

 

Hello,

It is NOT nice to remove the question.
That way others with the same problem don't have a clue of this thread.
This forum is to learn from each other.

Bertus