4) How do we obtain the expression of open loop undamped natural frequency response, w_{no} as (sqrt(2) * w_{c})?

Let's make \$\omega_c = 1\$ for the following calculation, for clarity sake. If the DC feedback is 1, you have \$G = \frac{G_0}{1 + G_0}\$ If you solve for \$G_0\$, you easily get: \$\frac{1}{G_0} = \frac{1}{G} - 1\$. From the Butterworth hypothesys, you know \$\frac{1}{G} = 1 + 2s + 2s^2 + s^3\$, so \$\frac{1}{G_0} = 2s + 2s^2 + s^3\$. That is, \$G_0 = \frac{1}{2s + 2s^2 + s^3} = \frac{1}{2s\left(1 + s + s^2/2\right)}\$

Using arbitrary \$\omega_c\$ again, this is (2.6):

\$\displaystyle G_0 \ = \ \frac{1}{\left(\frac{2}{\omega_c}s\right)\left(1 + \left(\frac{1}{\omega_c}\right)s + \left(\frac{1}{2\omega_c^2}\right)s^2\right)}\$

Now, (2.5) is the general form of a three pole system: one pole must be real, the other two must be complex conjugate. If you equate the general form (2.5) to the Butterworth form (2.6), the irreducible quadratic polynomials must be equal (proportional at least) , so:

\$\displaystyle 1 + \left(\frac{1}{\omega_c}\right)s + \left(\frac{1}{2\omega_c^2}\right)s^2 \ = \ 1 + 2\zeta_0\left(\frac{1}{\omega_{no}}\right)s + \left(\frac{1}{\omega_{no}^2}\right)s^2\$

Equating coefficients, you get: \$\frac{1}{\omega_c} = 2\zeta_0\frac{1}{\omega_{no}}\$ and \$\frac{1}{2\omega_c^2} = \frac{1}{\omega_{no}^2}\$

The second equation gives you \$\omega_{no} = \sqrt{2}\omega_c\$. Using this in the first equation gives you \$\zeta_0 = \frac{1}{\sqrt{2}}\$.