Calculating characteristic impedance of a matching line
You're required to find the \$ Zc \$ that will match the load to the line with a quarter wave transformer.

Because \$ R_g = Z_0 = 100 \Omega \$, the impedance at the junction, looking towards the generator, is \$ 100 \Omega \$. So you need the same impedance (actually, its complex conjugate) looking towards the load from the junction, i.e. \$ Z_{in} = 100 \Omega \$.

The load is fixed at \$ R_L = 400 \Omega \$, thus:

$$
Z_s = \sqrt{Z_{in} R_L} = 200 \Omega
$$

Now that we have matched the load through the quarter wave transformer, we can calculate the voltage phasor \$ V_L \$, because all the power available from the generator will be delivered to the load.

The maximum power available from the generator is:

$$
P_{avail} = \frac {1}{8} \frac {|V_g|^2} {R_g}
$$

And the power delivered to the load can be expressed as:

$$
P_L = \frac {1}{2} \frac {|V_L|^2} {R_L}
$$

Thus:

$$
\frac {1}{8} \frac {|V_g|^2} {R_g} = \frac {1}{2} \frac {|V_L|^2} {R_L}
$$

So we obtain the magnitude of the phasor:

$$
|V_L| = \frac {1}{2} |V_g| \sqrt {\frac{R_L}{R_g}} = 100V
$$

The phase is related to the electrical lengths of the transmission line sections:

$$
\arg(V_L) = - \beta_1 l_1 - \beta_s l_s = - \frac {2\pi}{\lambda_1} 1.2 \lambda_1 - \frac {2\pi}{\lambda_s} 0.25 \lambda_s = - 2.9 \pi
$$

And finally:

$$
V_L = 100V \cdot e^{- j 2.9 \pi}
$$