How is this transfer function generated using this LTIC circuit diagram?
For example: The third system with positive feedback - let's call \$H(s)\$ the resulting system:

$$ \frac{Y(s)}{F(s)} = H(s) = \frac{\omega_c}{s+0.1\omega_c} $$

Putting the pole in the standard form \$\frac{s}{\omega}+1\$ (in order to highlight the pole frequency):

$$ H(s) = \frac{10}{\frac{s}{\frac{\omega_c}{10}}+1} $$

So, the -3 dB frequency for this first order system is \$\omega_0=\frac{\omega_c}{10}\$ rad/s.

Alternatively, there is a long form to derive this result. Begin doing \$s = j\omega\$:

$$ H(j\omega)= \frac{10}{\frac{j\omega}{\frac{\omega_c}{10}}+1} $$

The module is:

$$ \left |H(j\omega) \right |= \frac{10}{\sqrt{\left ( \frac{10\omega}{\omega_c} \right )^2 +1}} $$

If \$\omega_0\$ is the -3dB cutoff frequency (in other words, the frequency when the amplitude reaches 0.707 of maximum one), we can write \$\left | H(j\omega_0) \right |= \frac{1}{\sqrt{2}}\left | H(0) \right |\$. 

As \$\left | H(0) \right | = 10\$:  

$$\frac{\left |H(j\omega_0)\right |}{\left | H(0) \right |} =  \frac{1}{\sqrt{\left ( \frac{10\omega_0}{\omega_c} \right )^2 +1}} = \frac{1}{\sqrt{2}}  $$ 

Resolving for \$\omega_0\$:

$$ \omega_0 = \frac{\omega_c}{10} $$ 

in rad/s.