Q1:

For the given circuit, find V_TH for the circuit external to R_L.

ANS:B - 4.6938.7° V

It is very simple.
1) As we know voltage across parallel is same so XL is useless to calculate;
2) To find vth RL is open circuit so remove RL.
3) Now we have VS= 6 V and Xc= 12 K is in series with R = 15 K.
4) Z = (R^2+Xc^2)^1/2 = 19.2 K.
5) Vth = R/Z * Vs = .78*6 = 4.69.
6) θ = tan-1 ( Xc/R) = .6747 rad * 57 = 38.5°.