Reactive power would put no extra load on a generator shaft if everything were perfect. However, real generators have real losses, with some of those proportional to the square of the current. The reactive load causes more current in the wires than there would be with a purely resistive load of the same real power. The extra current causes additional real power to be lost.
So the answer is that the engine will see a somewhat higher load and therefore use a bit more fuel. This is because of more inefficiencies and losses in the system, not be reactive power itself makes the generator harder to turn.
Added:
I should have mentioned this before, but somehow it slipped thru my mind at the time.
A reactive load on a perfect generator does not require more shaft power averaged over a cycle, but it does add "bumps" to the torque. One attribute of a 3 phase AC generator is that the torque is constant over a cycle with a resistive load. However, with a reactive load parts of the cycle will require more power and other parts less. The average power is still the same, but constant pushing forwards and backwards relative to the average torque can cause undesirable mechanical stresses and vibrations.
You can think of this a bit like moving two magnets past each other. Let's say they are oriented to repell. At a distance there is little force. You have to apply force to move them close together, meaning you put energy into the system. The magnets push in the direction of motion as they move away, thereby giving you back the energy you put in earlier. The net energy spent is 0, but there was definitely energy flow back and forth. There is always some loss as energy is moved around or converted back and forth in real systems.
Again, the reactive power itself doesn't cause the problem, but real power is lost because energy can't be moved around and converted with perfect efficiency. This real power loss has to be made up with more real power input. In addition, the extra mechanical forces can decrease the life of the generator and engine driving it.
Your total impedance is correct. As you can see cos(-19.653)=0.94 which is below 0.95.
If you calculate the total impedance when the 10 Ohms resistor is shorted you'll find 10<0, i.e. PF=1.
Therefore you can increase the PF by placing a small impedance accross the 10 Ohms resistor, and then the equivalent impedance modulus of both components will be below 10 Ohms (and below the small impedance). To get PF=1 you would place a really small reactance, whether it is an inductor or a capacitor.
To get PF=0.95 you can also succeed either with an inductor or with a capacitor.
Let's solve this using a capacitor. One way to do it is to calculate the input impedance (seen from the source).
If I is the current flowing from the source, the source voltage can be written as
$$ V=-j10·I + \frac { I·(j20+Zx) } {20 +j20+Zx }·20$$
The right term is the voltage in resistor 20 obtained by multiplying the current in this resistor (current divisor formula) by 20. And Zx is parallel between 10 Ohms and unknown reactance \$Zx=\frac {-jXc·10} {10-jXc}\$.
Therefore the Z, input impedance is
$$ Z=-j10 + \frac { (j20+\frac {-jXc·10} {10-jXc}) } {20 +j20+\frac {-jXc·10} {10-jXc} }·20$$
Using algebra (or some symbolic math package) you can get real part and imaginary part of Z.
Dividing Imaginary part and real part, and simplifying, you get:
$$\frac {\Im (Z)}{\Re{(Z)}}=-2.5·\frac {Xc^2} {7Xc^2-40Xc+400}$$
This division must be equal to $$-\tan(\arccos(0.95)\approx -0.329$$
Solving for Xc, you get 2 solutions, taking the positive one, \$Xc=8.821 \Omega \$. The capacitance is \$C \approx 361 \mu F\$
Best Answer
Ther mistake you made was assuming Q meant "reactive power" - it doesn't it's the Q-factor of the coil and quite simply is \$X_L/R\$. Q-factor is something you should look up. Q is sometimes called quality factor. See this wiki page. It's embodied in an RLC resonant tuned circuit for instance by this formula: -
And, for just the inductor on its own is simply \$X_L/R\$