To find the solution for the given system of equations:
\[
\begin{cases}
(x + 1)(4y + 1) = (2x - 1)(2y + 1) \\
(2x - 1)(3y + 1) = (3x - 2)(2y - 1)
\end{cases}
\]

We can substitute each set of values from the given options and check them.

Let's consider each option.

1. \((\frac{20}{31}, \frac{7}{31})\)

Substitute \(x = \frac{20}{31}\) and \(y = \frac{7}{31}\) into the first equation:
\[
\left(\frac{20}{31} + 1\right)\left(4 \cdot \frac{7}{31} + 1\right) = \left(2 \cdot \frac{20}{31} - 1\right)\left(2 \cdot \frac{7}{31} + 1\right)
\]
\[
\left(\frac{20}{31} + \frac{31}{31}\right)\left(\frac{28}{31} + \frac{31}{31}\right) = \left(\frac{40}{31} - \frac{31}{31}\right)\left(\frac{14}{31} + \frac{31}{31}\right)
\]
\[
\left(\frac{51}{31}\right)\left(\frac{59}{31}\right) = \left(\frac{9}{31}\right)\left(\frac{45}{31}\right)
\]
\[
\frac{3009}{961} \neq \frac{405}{961}
\]

Thus, (1) is not a solution. We follow the same substitution and verification method for the remaining options.

After verifying each option, we find that the correct solution is option (3):
\[
\left( \frac{20}{31}, - \frac{7}{31} \right)
\]

Thus, the solution to the given system of equations is:
\[
\left( \frac{20}{31}, - \frac{7}{31} \right)
\]