Homework questions with Compensated Attenuator
I don't want to answer these home work questions but I can show you how to determine the transfer function of this network in a swift and effortless manner. For this purpose, I will use the FACTs. The principle lies in determining the time constants of a given circuit in two different conditions: zeroed excitation (the input source is reduced to 0 V and replaced by a short circuit in your case) and nulled output (the excitation is back in place but does reach the output, leading to a 0 in the response). First, we will determine the gain (or attenuation in this resistive case) when \$s=0\$. To do that, open the caps as shown below:



The gain is simply \$H_0=\frac{R_2}{R_1+R_2}\$

Then, to obtain the time constant as indicated in your text, we reduce the excitation source \$V_{in}\$ to 0 V and replace it by a short. Then, we "look" through the individual capacitors terminal and determine the resistance:



But here, you see that both capacitors are in parallel. Despite the presence of two energy-storing elements, this is a 1st-order circuit with a single time constant. This is what is called a degenerate case in which the state variables of \$C_1\$ and \$C_2\$ are not independent but included in a loop with \$V_{in}\$: you loose one order in the denominator degree. The time constant in this case is \$\tau_2=(C_1+C_2)(R_1||R_2)\$. The pole in a 1st-order circuit is the inverse of the time constant: \$\omega_p=\frac{1}{\tau_2}=\frac{1}{(C_1+C_2)(R_1||R_2)}\$.

The zero is obtained by looking at an impedance condition in this circuit in which the excitation \$V_{in}\$ would not produce a response \$V_{out}\$.



This is when the impedance made of \$C_1\$ paralleled with \$R_1\$ becomes infinite. Otherwise stated, \$Z_1(s)=\frac{N(s)}{1+sR_1C_1}=0\$ when \$s_z=-\frac{1}{R_1C_1}\$ or \$\omega_z=\frac{1}{R_1C_1}\$. This is it, the complete transfer function is expressed as \$H(s)=H_0\frac{1+\frac{1}{\omega_z}}{1+\frac{1}{\omega_p}}\$. The below Mathcad sheet shows the expression at work and compares the results with the brute-force expression in the upper right corner:



I have purposely selected two different time constants but the principle is to match them to obtain flat magnitude and phase responses. You can see thanks to the low-entropy factorization that \$H_0\$ sets the dc attenuation (but if the 1-Mohm scope input impedance comes in || with \$R_2\$, it slightly changes the picture) but also the output resistance. The FACTs are unbeatable in terms of efficiency and ease of implementation. You saw how I could obtain this transfer function via three simple sketches that I could later individually fix in case of a typo.