Given circuit is

Applying KVL to the mesh1 we have
................. (1)
Applying KVL to the super mesh we have
................ (2)
From the circuit
since
Substituting in equation(2)

Substituting in equation(1)

and
(b)
Power delivered in
is
Power genereated in voltage source is
Power genereated in dependent current source is
Voltage across the current source
Now power
Amplifier efficiency,