Answer: \(\mathrm P(X>2)=\boxed{0.449}\), and the interquartile range is \(\boxed{2.75}\).

(i) For \(x\geq0\),

\[F(x)=1-e^{-0.4x}.\]

Therefore

\[\mathrm P(X>2)=1-F(2).\]

Now

\[F(2)=1-e^{-0.8}.\]

So

\[\mathrm P(X>2)=e^{-0.8}=0.449\quad\text{to 3 significant figures}.\]

(ii) The lower quartile \(Q_1\) satisfies \(F(Q_1)=0.25\). Thus

\[1-e^{-0.4Q_1}=0.25.\]

So

\[e^{-0.4Q_1}=0.75=\frac34,\]

and hence

\[Q_1=\frac{\ln(4/3)}{0.4}=0.719\ldots.\]

The upper quartile \(Q_3\) satisfies \(F(Q_3)=0.75\). Therefore

\[1-e^{-0.4Q_3}=0.75,\]

so

\[e^{-0.4Q_3}=0.25=\frac14.\]

Hence

\[Q_3=\frac{\ln4}{0.4}=3.466\ldots.\]

The interquartile range is

\[Q_3-Q_1=3.466\ldots-0.719\ldots=2.746\ldots.\]

So the interquartile range is \(\boxed{2.75}\) to 3 significant figures.