Non-inverting: zero at -4, pole at -8, HF gain 2

Required function: zero at $s=-4$, pole at $s=-8$, HF gain 2:

$$T(s) = 2\,\frac{s+4}{s+8}$$

DC gain $= 2 \times 4/8 = 1$. The zero is closer to the origin than the pole (a lead-type network): the gain rises from 1 at DC to 2 at high frequency, with phase lead in between.

Non-inverting configuration: $Z_1$ from the inverting input to ground, $Z_2$ from output to inverting input:

$$T(s) = 1 + \frac{Z_2}{Z_1}$$

Use $Z_2 = R_2$ and $Z_1 = R_1 + \dfrac{1}{sC_1}$ (series RC to ground):

$$\begin{aligned} T(s) &= 1 + \frac{sR_2C_1}{1 + sR_1C_1} = \frac{1 + s(R_1+R_2)C_1}{1 + sR_1C_1} \\ &= \frac{R_1 + R_2}{R_1}\cdot\frac{s + \frac{1}{(R_1+R_2)C_1}}{s + \frac{1}{R_1C_1}} \end{aligned}$$

              +---R2---+
              |        |
   +----------+-|-\    |
   |            |  >---+--- Vo
  R1    Vin-----|+/
   |
  === C1
   |
  GND

Matching:

$$\begin{aligned} \frac{1}{R_1C_1} &= 8 \\ \frac{1}{(R_1 + R_2)C_1} &= 4 \Rightarrow R_1 + R_2 = 2R_1 \Rightarrow R_2 = R_1 \\ \text{HF gain} &= \frac{R_1+R_2}{R_1} = 2 \;(\text{satisfied}) \end{aligned}$$

Normalized ($C_1 = 1$ F): $R_1 = R_2 = 0.125\ \Omega$.

Practical values: choose $C_1 = 10\ \mu$F: $R_1 = \dfrac{1}{8 \times 10^{-5}} = 12.5\ \text{k}\Omega$, $R_2 = 12.5\ \text{k}\Omega$.

Answer: $R_1 = R_2 = 12.5\ \text{k}\Omega$, $C_1 = 10\ \mu$F; $T(s) = 2(s+4)/(s+8)$.