Stop speaking French to me 😡😡😡😡redo question 6 in English

Initial activity $A_0 = 1.2 \times 10^5 \text{ Bq}$.  
Activity after $t = 12 \text{ days}$ is $A = 1.5 \times 10^4 \text{ Bq}$.
$$
\frac{A}{A_0} = \left(\frac{1}{2}\right)^n
$$
$$
\frac{1.5 \times 10^4 \text{ Bq}}{1.2 \times 10^5 \text{ Bq}} = \left(\frac{1}{2}\right)^n
$$
$$
0.125 = \left(\frac{1}{2}\right)^n
$$

We know that $0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3$.  
Therefore, $n = 3$.

$$
3 = \frac{12 \text{ days}}{T_{1/2}}
$$
$$
T_{1/2} = \frac{12 \text{ days}}{3}
$$
$$
T_{1/2} = 4 \text{ days}
$$
The half-life of the isotope is $\boxed{4 \text{ days}}$.

Total time $t_{\text{total}} = 12 \text{ days} + 20 \text{ days} = 32 \text{ days}$.

$$
n' = \frac{t_{\text{total}}}{T_{1/2}} = \frac{32 \text{ days}}{4 \text{ days}}
$$
$$
n' = 8
$$

$$
A' = A_0 \left(\frac{1}{2}\right)^{n'}
$$
$$
A' = (1.2 \times 10^5 \text{ Bq}) \times \left(\frac{1}{2}\right)^8
$$
$$
A' = (1.2 \times 10^5 \text{ Bq}) \times \frac{1}{256}
$$
$$
A' \approx 468.75 \text{ Bq}
$$
$$
A' \approx 4.7 \times 10^2 \text{ Bq}
$$
The activity of the sample after an additional 20 days will be approximately $\boxed{4.7 \times 10^2 \text{ Bq}}$.