Answer
(a) 15.8 years
(b) 0.88 or 88%
Work Step by Step
(a) $\frac{1}{8}=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$
The time required is three half-lives.
That is, $t=3\times5.27\,y=15.8\,years$
(b) Half-life $t_{1/2}=5.27\,y$
Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{5.27\,y}=0.1315\,y^{-1}$
$t=1.0\,y$
Recall that $\ln(\frac{A}{A_{0}})=-\lambda t$ where $A_{0}$ is the activity of sample at the beginning and $A$ is the activity after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(0.1315\,y^{-1})(1.0\,y)=-0.1315$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-0.1315}=0.88$
