Answer: \(F(x)=0\) for \(x<0\), \(F(x)=1-e^{-0.2x}\) for \(x\ge 0\); \(\boxed{P(X>2)=0.670}\); median \(\boxed{5\ln2=3.47}\).

(i) For \(x<0\), the density is zero, so

\[F(x)=0.\]

For \(x\ge 0\),

\[F(x)=\int_0^x 0.2e^{-0.2t}\,dt.\]

Since

\[\int 0.2e^{-0.2t}\,dt=-e^{-0.2t},\]

we get

\[F(x)=\left[-e^{-0.2t}\right]_0^x.\]

Therefore

\[F(x)=-e^{-0.2x}+1=1-e^{-0.2x}.\]

So

\[\boxed{F(x)=\begin{cases}0, & x<0,\\[2mm]1-e^{-0.2x}, & x\ge0.\end{cases}}\]

(ii) We want the upper tail probability:

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

Thus

\[P(X>2)=e^{-0.2(2)}=e^{-0.4}.\]

Hence

\[P(X>2)=\boxed{0.670}\quad\text{to 3 significant figures}.\]

(iii) Let the median be \(m\). Then

\[F(m)=\frac12.\]

So

\[1-e^{-0.2m}=\frac12.\]

Hence

\[e^{-0.2m}=\frac12.\]

Taking logarithms,

\[-0.2m=\ln\left(\frac12\right)=-\ln2.\]

Therefore

\[m=\frac{\ln2}{0.2}=5\ln2.\]

So the median is

\[\boxed{5\ln2=3.47}\quad\text{to 3 significant figures}.\]