Calculate transfer function given 3dB point

Let me label your circuit nodes for better understanding of your question

^{simulate this circuit – Schematic created using CircuitLab}

To answer your question I must first clarify some definitions:

• Error gain function: a gain k that multiplies the error e = r-y
• Transfer function: the behavior of the closed-loop system y/r

Now,

$$ y=k\cdot e $$ $$ e=r-y $$ $$ y=k\cdot (r-y) $$

So, the error gain function follows as

$$ k=\frac{y}{r-y} $$

So, you don't even need the transfer function to establish the value of k. You just need y and r.

Anyway the transfer function is

$$ TF=\frac{y}{r}=\frac{k}{k+1} $$

So, if you're given the transfer function (note that this is just a constant for this theoretical system) you can find k as

$$ k=\frac{TF}{1-TF} $$

Note also that, with negative feedback, in stable system conditions, and with finite k:

$$ 0\leq TF<1 $$

Using impedances (forgive the lack of standard forms along the way) and going long-hand from scratch, I get:

$$\begin{align*} \frac{V_O}{V_I}&= \frac{R\:\vert\vert\: C_2}{C_1+R\:\vert\vert\: C_2}\\\\ &=\frac{\frac{R}{1+s R C_2}}{\frac{1}{s C_1}+\frac{R}{1+s R C_2}}\\\\&=\frac{\frac{R}{1+s R C_2}}{\frac{1+s R C_2}{s C_1 \left(1+s R C_2\right)}+\frac{s R C_1}{s C_1 \left(1+s R C_2\right)}}\\\\ &=\frac{\frac{R}{1+s R C_2}}{\frac{1+s R\cdot\left(C_1+C_2\right)}{s C_1 \left(1+s R C_2\right)}}=\frac{R}{1+s R C_2}\cdot\frac{s C_1 \left(1+s R C_2\right)}{1+s R\cdot\left(C_1+C_2\right)}\\\\&=\frac{s R C_1}{1+s R\cdot\left(C_1+C_2\right)}=\frac{\frac{s C_1}{C_1+C_2}}{s+\frac{1}{R\cdot\left(C_1+C_2\right)}}\\\\&=\left[\frac{s}{s+\frac{1}{R\cdot\left(C_1+C_2\right)}}\right]\cdot\left[\frac{C_1}{C_1+C_2}\right] \end{align*}$$

So I guess I agree with your results (in the first part.)