Answer
(a) $5$ days
(b) $18.96\%$ of original mass left.
Work Step by Step
(a) Given $m(8)=0.33m_0$, use the radioactive decay model $m(t)=m_0e^{-rt}$, we have
$0.33m_0=m_0e^{-8r}$ which gives $r=-ln(0.33)/8$ and half-life $h=ln2/r=-8ln2/ln(0.33)\approx5$ days
(b) Let $t=12$, we have $m(12)=m_0e^{-12ln2/5}\approx0.1895m_0$ which mean $18.96\%$ of original mass left.
