(i) the absorption of energy/loss of power from the beam;

(ii) correct substitution $\frac{{{I_0}}}{2} = {I_0}{e^{ - \mu {x_{\frac{1}{2}}}}}$;
taking natural logs $1{\rm{n}}\frac{1}{2} =  - \mu {x_{\frac{1}{2}}}$;
$\left( {1{\rm{n}}2 = \mu {x_{\frac{1}{2}}}} \right)$
Answer given, marks are for correct working.

(iii) $\mu  = \left( {\frac{{1{\rm{n}}2}}{{0.73}} = } \right)0.95{\rm{c}}{{\rm{m}}^{ - 1}}$;
$I = \left( {{I_0}{e^{ - 0.95 \times 2.0}} = } \right)0.15{I_0}$;

or

number of half-value thicknesses $= \frac{2}{{0.73}} = 2.74$;
$I = {0.5^{2.74}} = 0.15{I_0}$;
Award [2] for a bald correct answer.