Is the change in Kinetic Energy not equal to the work done when mass changes at some rate? 
Here the Rate of change in Kinetic energy is not equal to the the power. Please explain
 A: Official answer to point (ii) is not correct. Proof (counterexample):
Note: I will write as $V$ uppercase the final water speed.
In an interval $T$ an amount of water $\rho A V T$ must be accelerated from $0$ to $V$. Assuming it is done with a constant acceleration $V/T$ then:
$$F=ma=(\rho A V T) (V/T)=\rho A V^2$$
$$displacement = x = \frac{1}{2} (V/T) T^2=\frac{1}{2} VT$$ (here the official answer made a mistake, it fix $x=VT$)
$$work = W = Fx = (\rho A V^2) (\frac{1}{2} VT) = \frac{1}{2} \rho A V^3 T$$
$$power=W/T=\frac{1}{2} \rho A V^3$$
With these results, there are not mismatch between the calculus by energy and by force.
A: Actually, I must disagree with @pasaba por aqui's answer.  The textbook answer given in the OP is correct - the power provided is twice the gain in KE of the stream.  The other half of the power is dissipated.  This is sometimes known as the "conveyor belt paradox," because a similar setup where it arises is the case of sand being deposited with zero speed onto a conveyor belt moving at constant speed $\text{v}$.  The force that must be applied to the belt is $\dot{m}\text{v}$, and the power to drive the belt is $\dot{m}\text{v}^2$, while the rate of increase of KE of the sand is only $\frac{1}{2}\dot{m}\text{v}^2$.  For a good explanation of why half the power gets dissipated (at least in the case of the sand), if you have access, see this article   (F. S. Crawford, The Physics Teacher 27, 546 (1989)).
