Answer
a) $1.63\%$
b) $0.6631\%$
Work Step by Step
a) Let's assume that the sample amount is $m$ since the author did not give us a definite mass of the sample.
So,
$$m_{{235}}=\dfrac{0.72}{100}m=\color{blue}{0.0072}\;m$$
and
$$m_{ {238} }=\left[1-\dfrac{0.72}{100}\right]m=\color{blue}{0.9928}\; m$$
Now we need to find the ratio between two amounts 1.0 billion years ago.
We know that
$$N=N_0e^{-\lambda t}$$
where $N$ is for today and $N_0$ is for 1.0 billion years ago where both of them are for $\rm ^{235}_{92}U$.
Solving for $N_0$;
$$\dfrac{N_{235}}{e^{-\lambda t}}=N_{0,235}$$
$$ N_{0,235}=e^{\lambda t} N_{235} $$
Recalling that $\lambda=\dfrac{\ln 2}{T_{\frac{1}{2}}}$;
$$ N_{0,235}=e^{\frac{(\ln 2)t}{T_{\frac{1}{2}}}} N_{235} $$
Plugging the known;
$$ N_{0,235}=e^{\frac{ (\ln 2) (1\times 10^9)}{ (7.04\times 10^8)} }\times0.0072m=\color{blue}{0.01927}m $$
By the same approach;
$$ N_{0,238}=e^{\frac{ (\ln 2) (1\times 10^9)}{ (4.468\times 10^9)} }\times 0.9928m=\color{blue}{1.1594}m $$
Now we can easily find the percentage of $\rm ^{235}_{92}U$ by
$${\rm ^{235}_{92}U\%}=\dfrac{N_{0,235}}{N_{0,238}+N_{0,235}}\times 100\%$$
$${\rm ^{235}_{92}U\%}=\dfrac{0.01927m}{1.1594m+0.01927m}\times 100\%=\color{red}{\bf1.63\%}$$
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b) Applying the same approach as we did above but for $10^8$ years in the future.
We need to use the amount of today as $N_0$.
$$N_{235}=N_{0,235}e^{-\lambda t}$$
$$N_{235}=N_{0,235}e^{\left[-\frac{\ln 2}{T_{\frac{1}{2}}}t\right]}$$
Plugging the known;
$$ N_{235}=e^{-\frac{ (\ln 2) (1\times 10^8)}{ (7.04\times 10^8)} }\times 0.0072m=\color{blue}{6.5249\times10^{-3}}m $$
By the same approach;
$$ N_{238}=e^{-\frac{ (\ln 2) (1\times 10^8)}{ (4.468\times 10^9)} }\times0.9928 m=\color{blue}{977.52\times10^{-3}}m $$
Hence;
$${\rm ^{235}_{92}U\%}=\dfrac{N_{0,235}}{N_{0,238}+N_{0,235}}\times 100\%$$
$${\rm ^{235}_{92}U\%}=\dfrac{6.5249\times10^{-3}m}{977.52\times10^{-3}m+6.5249\times10^{-3}m}\times 100\%$$
$${\rm ^{235}_{92}U\%}=\color{red}{\bf 0.6631\%}$$