Answer
(a).$h=3.82$
(b).$t=8.87$
Work Step by Step
$m(t)=m_0e^{-rt}$. Whereas,$m(t)$ is the mass of radioactive substance after time $t$, $m_0$ is the Initial mass of radioactive substance, $r=\frac{\ln 2}{h}$ is the rate of decay while $h$ is the half-life and $t$ is time.
$m(t)=m_02^{-t/h}$. Whereas,$m(t)$ is the mass of radioactive substance after time $t$, $m_0$ is the Initial mass of radioactive substance, $h$ is the half-life and $t$ is time.
(a).$t$ is 3 days, $m_0=1$, $m(t)=0.58$
$\frac{m(t)}{m_0}=2^{-t/h}$,
$\log (\frac{m(t)}{m_0})=-t/h \log2$,
$h=\frac{-\log2 t}{\log(m(t)/m_0)}=3.82$ days
(b).$h=3.82$, $m_0=1$, $m(t)=0.2$
$\frac{m(t)}{m_0}=2^{-t/h}$,
$\log (\frac{m(t)}{m_0})=-t/h \log2$,
$t=\frac{-\log(m(t)/m_0)h}{\log2}=8.87$ days