Formula derivation of Doherty power amplifier from circuit diagram
My results don't agree with what you show. The results would match though if \$R_L=Z_W\$. But here is what I did:
Question 1
Let's turn off (open circuit) current source \$I_S\$, then, forward current in the T-line:
$$I^+ = I_H$$
Reflection coefficient at the load is:
$$\Gamma_L = \frac{R_L - Z_W}{R_L+Z_W}$$
Thus the current into the load \$R_L\$ is:
$$I_L = (1-\Gamma_L)I^+e^{-j\frac{2\pi}{\lambda}.\frac{\lambda}{4}} = -\frac{2Z_W}{R_L+Z_W}jI_H$$
$$V_{L1} = I_LR_L = -\frac{2Z_WR_L}{R_L+Z_W}jI_H$$
Now turn off the current source \$I_H\$ and the T-line acts as a short circuit, so 
$$V_{L2} = 0$$
Thus, using superposition,
$$V_{L} = -\frac{2Z_WR_L}{R_L+Z_W}jI_H$$

Question 2
Turn off the current source \$I_S\$, and the impedance looking into the line is:
$$Z_L = \frac{Z_W^2}{R_L}$$
Thus,
$$V_{H1} = \frac{Z_W^2}{R_L}I_H$$
Now, turn off current source \$I_H\$, the forward current in the line is:
$$I^+ = -\frac{R_L}{R_L+Z_W}jI_S$$
Thus, 
$$V^+ = I^+Z_W = -\frac{R_LZ_W}{R_L+Z_W}jI_S$$
The reflection coefficient at the load is now unity, thus total voltage at load is:
$$V_{H2} = (1+1)V^+e^{-j\frac{2\pi}{\lambda}.\frac{\lambda}{4}} = -\frac{2R_LZ_W}{R_L+Z_W}I_S$$
Thus, total voltage using superposition:
$$V_H = \frac{Z_W^2}{R_L}I_H-\frac{2R_LZ_W}{R_L+Z_W}I_S$$

Question 3
The impedance \$Z_S\$ is, by definition:
$$Z_S = \frac{V_L}{I_S}$$
$$Z_S = \frac{2Z_WR_L}{R_L+Z_W}\frac{I_H}{I_S}$$