Solve the system of equations: { (2x+3)/(3y-2) = 1; x(2y-5)-2y(x+3) = 2x+1; AND { (x+1)/(y+2) = 5; 3(2x-5)-4(3y+4) = 5.

This problem contains two independent systems of equations. Let's solve them one by one.

System 1:

\[ \begin{cases} \frac{2x+3}{3y-2} = 1 \\ x(2y-5) - 2y(x+3) = 2x+1 \end{cases} \]

1. Simplify the first equation:
   \[ 2x+3 = 3y-2 \]
   \[ 2x - 3y = -5 \]
2. Simplify the second equation:
   \[ 2xy - 5x - 2xy - 6y = 2x+1 \]
   \[ -5x - 6y = 2x+1 \]
   \[ -7x - 6y = 1 \]
3. Now we have a new system:
   \[ \begin{cases} 2x - 3y = -5 \\ -7x - 6y = 1 \end{cases} \]
4. Multiply the first equation by 2 to eliminate y:
   \[ 4x - 6y = -10 \]
5. Subtract the new first equation from the second equation:
   \[ (-7x - 6y) - (4x - 6y) = 1 - (-10) \]
   \[ -7x - 4x - 6y + 6y = 1 + 10 \]
   \[ -11x = 11 \]
   \[ x = -1 \]
6. Substitute x = -1 into the simplified first equation (2x - 3y = -5):
   \[ 2(-1) - 3y = -5 \]
   \[ -2 - 3y = -5 \]
   \[ -3y = -3 \]
   \[ y = 1 \]

Solution for System 1: x = -1, y = 1

System 2:

\[ \begin{cases} \frac{x+1}{y+2} = 5 \\ 3(2x-5) - 4(3y+4) = 5 \end{cases} \]

7. Simplify the first equation:
   \[ x+1 = 5(y+2) \]
   \[ x+1 = 5y+10 \]
   \[ x - 5y = 9 \]
8. Simplify the second equation:
   \[ 6x - 15 - 12y - 16 = 5 \]
   \[ 6x - 12y - 31 = 5 \]
   \[ 6x - 12y = 36 \]
9. Divide the second equation by 6:
   \[ x - 2y = 6 \]
10. Now we have a new system:
    \[ \begin{cases} x - 5y = 9 \\ x - 2y = 6 \end{cases} \]
11. Subtract the second equation from the first equation:
    \[ (x - 5y) - (x - 2y) = 9 - 6 \]
    \[ x - x - 5y + 2y = 3 \]
    \[ -3y = 3 \]
    \[ y = -1 \]
12. Substitute y = -1 into the simplified second equation (x - 2y = 6):
    \[ x - 2(-1) = 6 \]
    \[ x + 2 = 6 \]
    \[ x = 4 \]

Solution for System 2: x = 4, y = -1

Final Answer:

System 1: x = -1, y = 1

System 2: x = 4, y = -1