Fourth-order bandpass filter Bessel - I can not solve the final part (math accounts)
If your system of equations is

\$G_1(G_4+G_5)C_2= k_1\$

\$G_2G_3G_5= k_2\$

\$\frac{G_1G_4C_2}{G_2G_3G_5}= k_3\$

\$\frac{G_4C_1C_2}{G_2G_3G_5}= k_4\$

and you want to pre-assign \$C_1\$ and \$C_2\$ then you are not free to choose any arbitrary other variables to assign a value by yourself but only particular ones.

You have to find out which ones:

\$k_3/k_4 = G_1 / C_1\$, \$\Rightarrow\$ \$G_1\$ is already determined.

\$k_2 k_3 = G_1 G_4 C_2\$, \$\Rightarrow\$ also \$G_4\$ is already determined.

\$k_2 k_4 = G_4 C_1 C_2\$, \$\Rightarrow\$ also \$G_4\$ is already determined (no new Information)

Now looking at 1st equation shows that also \$G_5\$ is determined (because \$G_1, G_4\$ and \$C_2\$ are determined).

Finally note that \$G_2\$ and \$G_3\$ appear only as product \$G_2 G_3\$ in any of the equations, i.e. you can choose one of them without affecting the solubility of the system.

Conclusion:
If you pre-determine \$C_1\$ and \$C_2\$ the system of equations also determines \$G_1, G_4\$ and \$G_5\$; i.e. you are not free to choose any of those \$G\$'s.

You should, however, be able to solve the system of equations by
assigning values to either


\$C_1, C_2\$ and \$G_2\$ or
\$C_1, C_2\$ and \$G_3\$


And if you don't want to do it manually (or if you want to verify the result you found manually) this is how you can use sympy to do the algebra:

from sympy import *

# Define symbols
G1, G2, G3, G4, G5 = symbols('G1 G2 G3 G4 G5')
C1, C2 = symbols('C1 C2')
k1, k2, k3, k4 = symbols('k1 k2 k3 k4')

# Solve equation system for for Vout. 
solvedEqSystem = solve([ 
        G1 * (G4 + G5) * C2           - k1, # equation (1)
        G2 * G3 * G5                  - k2, # equation (2)
        C2 * G1 * G4 / (G2 * G3 * G5) - k3, # equation (3)
        C1 * C2 * G4 / (G2 * G3 * G5) - k4  # equation (4)
    ],
    # list of variables to solve for; i.e. all the ones not listed here 
    # (C1, C2, G3) are assumed to be pre-detemined and will remain in 
    # the resulting expressions
    [G1, G2, G4, G5]  
)

for r in solvedEqSystem[0]:
    print r


Result: 4 expressions for the 4 unknon variables \$G_1, G_2, G_4, G_5\$:

C1*k3/k4
C1*C2*k2*k3/(G3*k4*(k1 - k2*k3))
k2*k4/(C1*C2)
k4*(k1 - k2*k3)/(C1*C2*k3)