Power factor correction in a simple phasor circuit
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&lt;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\$