Consider the reversible processes for 1.0 mol of an ideal gas as...

Using the IUPAC sign convention for thermodynamics, work done is given by $w = -\int p dV$.

For the isothermal reversible process 1 (from volume $V_1$ to $V_2$ at constant temperature $T_1$):
$w_1 = -nRT_1 \ln\left(\dfrac{V_2}{V_1}\right)$
Given $n = 1$ mol and $R = 2$ cal K$^{-1}$ mol$^{-1}$, we have:
$w_1 = -2T_1 \ln\left(\dfrac{V_2}{V_1}\right)$

For the isothermal reversible process 3 (from volume $V_3$ to $V_4$ at constant temperature $T_2$):
$w_3 = -nRT_2 \ln\left(\dfrac{V_4}{V_3}\right) = -2T_2 \ln\left(\dfrac{V_4}{V_3}\right)$

Adding the work done in these two processes:
$w_1 + w_3 = -2T_1 \ln\left(\dfrac{V_2}{V_1}\right) - 2T_2 \ln\left(\dfrac{V_4}{V_3}\right)$

Let us also verify the other options to be sure.
For option (1): The net work done in a reversible cycle is the area enclosed by the $p-V$ curve, which is non-zero. Thus, $w_1 + w_2 + w_3 + w_4 \neq 0$.

For option (3): In adiabatic processes, $q = 0$, so from the first law of thermodynamics, $\Delta U = w$.
Therefore, $w_2 = \Delta U_2$ and $w_4 = \Delta U_4$.
This gives $w_2 + w_4 = \Delta U_2 + \Delta U_4$.
Since internal energy is a state function, for the complete cycle $\Delta U_{cycle} = \Delta U_1 + \Delta U_2 + \Delta U_3 + \Delta U_4 = 0$.
For isothermal processes 1 and 3, $\Delta U_1 = 0$ and $\Delta U_3 = 0$.
Thus, $\Delta U_2 + \Delta U_4 = 0$, which means $w_2 + w_4 = 0$.
However, $\Delta U_2 - \Delta U_4 = \Delta U_2 - (-\Delta U_2) = 2\Delta U_2 \neq 0$.
Hence, $w_2 + w_4 \neq \Delta U_2 - \Delta U_4$.

For option (4): $w_1 + w_2 = -2T_1 \ln\left(\dfrac{V_2}{V_1}\right) + \Delta U_2 \neq 2T_1 \ln\left(\dfrac{V_2}{V_1}\right)$.

Answer: $w_1+w_3=-2T_1\ln\dfrac{V_2}{V_1}-2T_2\ln\dfrac{V_4}{V_3}$