在一块石头中检测到的放射性物质的质量为 $m$ 克，由公式 $m = ae^{-kt}$ 给出，其中 $a \ne 0$，$k$ 是一个常数，$t$ 是以月为单位的时间间隔。

1. Set up the equation using the given condition for half-life

At $t = 0$, the original mass is $m = ae^0 = a$.
When $t = 4$, $m = 0.5a$.
$0.5a = ae^{-4k}$
$0.5 = e^{-4k}$

2. Solve for $k$ by taking the natural logarithm

$\ln 0.5 = -4k$
$k = \frac{\ln 0.5}{-4}$
$k = \frac{-\ln 2}{-4} = \frac{\ln 2}{4}$
$k \approx 0.1733 \text{ (to 4 s.f.)}$

1. Recall or calculate the value of $k$ from part (i)

From part (i), $0.5a = ae^{-4k} \implies e^{-4k} = 0.5 \implies k = -\frac{1}{4}\ln 0.5 = \frac{1}{4}\ln 2 \approx 0.17329$.

2. Substitute the given values into the model to find the initial mass $a$

Given $m = 0.25$ when $t = 1$:
$0.25 = ae^{-k(1)}$
$a = \frac{0.25}{e^{-0.17329}} = 0.25e^{0.17329}$
$a \approx 0.29730$

1. Recall the values of $k$ and $a$ from parts (i) and (ii) and substitute $m = 0.01$ into the model

From (i), $k = \frac{1}{4}\ln 2 \approx 0.17329$.
From (ii), $a = 0.25e^k = 0.25e^{\frac{1}{4}\ln 2} \approx 0.29730$.
Substitute $m = 0.01$ into $m = ae^{-kt}$:
$0.01 = 0.29730 e^{-0.17329t}$

2. Solve for $t$ using logarithms

$e^{-0.17329t} = \frac{0.01}{0.29730}$
$-0.17329t = \ln\left(\frac{0.01}{0.29730}\right)$
$-0.17329t \approx -3.39216$
$t \approx 19.575$
$t = 19.6 \text{ months (to 3 s.f.)}$