Natural water contains a trace amount of tritium (^ 3 _ 1 H ). This...

Let the initial radioactivity of the natural water used to make the whisky be $A_0$.

For the recently purchased bottle marked '8 years old', the water was drawn 8 years ago. Its present radioactivity is:
$A_1 = A_0 e^{-8\lambda}$

Assuming the found bottle was also an '8 years old' whisky when it was left on the peak $t$ years ago, the total time elapsed since its water was drawn is $(t + 8)$ years. Its present radioactivity is:
$A_2 = A_0 e^{-(t+8)\lambda}$

Given that the found bottle contains $1.5\%$ of the radioactivity of the recently purchased bottle:
$A_2 = 0.015 A_1$

Substituting the expressions for $A_1$ and $A_2$:
$A_0 e^{-(t+8)\lambda} = 0.015 A_0 e^{-8\lambda}$

Dividing both sides by $A_0 e^{-8\lambda}$:
$e^{-\lambda t} = 0.015$

Taking the natural logarithm on both sides:
$-\lambda t = \ln(0.015)$

Using the decay constant formula $\lambda = \dfrac{\ln 2}{T_{1/2}}$:
$t = \dfrac{-\ln(0.015)}{\ln 2} \times T_{1/2}$

$t = \dfrac{\ln\left(\dfrac{1000}{15}\right)}{\ln 2} \times 12.5$

$t = \dfrac{\ln(66.67)}{\ln 2} \times 12.5$

$t \approx 6.059 \times 12.5 \approx 75.7 \text{ years}$

The time elapsed since the unsuccessful attempt is approximately $75.7 \text{ years}$.