Answer
(a) 28.1 years
(b) 8.66 years
Work Step by Step
(a) Decay rate= $1238\,dpm=kN=\frac{0.693}{t_{1/2}}\times2.64\times10^{10}$
($k$ is the decay constant, $N$ is the number of particles)
$\implies$Half-life $t_{1/2}=\frac{0.693}{1238/min}\times2.64\times10^{10}$
$=1.4778\times10^{7}\,min=1.4778\times10^{7}\,min\times\frac{1\,h}{60\,min}\times\frac{1\,d}{24\,h}\times\frac{1\,y}{365\,d}$
$=28.1\,y$
(b) Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1\,y}=0.02466\,y^{-1}$
Original rate $R_{0}=1238\,dpm$
Rate after time $t$, $R=1000\,dpm$
Recall that $\ln(\frac{R_{0}}{R})=kt$, where $t$ is the time required.
$\implies \ln(\frac{1238\,dpm}{1000\,dpm})=0.2135=0.02466\,y^{-1}(t)$
$\implies t=\frac{0.2135}{0.02466\,y^{-1}}=8.66\,y$
