$t_{1/2} = 5730$ years

$t = 5340 + 1000 = 6340$ years (since the remains were found in year 2991 or 1000 years after first discovery in Example 26.6)

$R_{initial} = 0.231 Bq$

$textbf{Find:} R_{final}$

$textbf{Formula:}$

$$
begin{gather}
t = dfrac{t_{1/2}}{0.693} times ln dfrac{R_{initial}}{R_{final}}
end{gather}
$$

Using Eq (1), we can obtain an equation that will solve for $R_{final}$, such that:

$$
begin{align*}
6340 text{ years} &= dfrac{5730 text{ years}}{0.693} ln dfrac{0.231 text{ Bq}}{R_{final}} \
dfrac{(6340)(0.693)}{5730} &= ln dfrac{0.231}{R_{final}} \
0.7667 &= ln dfrac{0.231}{R_{final}} \
end{align*}
$$

We then raise both sides of the equation by $e$, as follows:

$$
begin{align*}
e^{0.7667}&= e^{left(ln dfrac{0.231}{R_{final}}right)} \
2.1526 &= dfrac{0.231}{R_{final}} \
R_{final} &= dfrac{0.231}{2.1526} \
R_{final} &= 0.1073 text{ Bq}
end{align*}
$$

Hence, the remaining carbon-14 activity of the remains will be $boxed{0.1073 text{ Bq}}$

$R_{final} = 0.185$ Bq

$textbf{Knowns:}$

$t_{1/2} = 5730$ years

$R_{initial} = 0.231$ Bq

$textbf{Formula:}$
$$
begin{gather}
t = dfrac{t_{1/2}}{0.693} ln left(dfrac{R_{initial}}{R_{final}} right)
end{gather}
$$

Using Eq (1), we can calculate for $t$ as follows:

$$
begin{align*}
t &= dfrac{5730}{0.693} ln left(dfrac{0.231}{0.185}right) \
t &= 1836
end{align*}
$$

Hence, the fire burned $boxed{text{1836 years ago}}$

half-life = $432$ y

$R_{final} = dfrac{R_{initial}}{525}$

Formula:

time = $dfrac{text{half-life}}{0.693} ln left (dfrac{R_{initial}}{R_{final}} right )$

To determine how long this kind of smoke detector will work, we can calculate the time using the formula above, such that:

$$
begin{align*}
time &= dfrac{432 text{y}}{0.693} ln left (dfrac{R_{initial}}{dfrac{R_{initial}}{525}} right ) \
&= dfrac{432 text{y}}{0.693} ln (525) \
&= 3094 y
end{align*}
$$

Hence, the smoke detector will work for $boxed{3904 text{ y}}$