Answer: \(\boxed{a=\ln2}\), \(\boxed{E(X)=\dfrac1{\ln2}}\), \(\boxed{\mathrm{IQR}=\dfrac{\ln3}{\ln2}}\), and \(\boxed{g(y)=\dfrac1{y^2}\text{ for }y\ge1}\).

(i) The total probability must be 1:

\[\int_0^\infty ae^{-x\ln2}\,dx=1.\]

Now

\[\int_0^\infty e^{-x\ln2}\,dx=\frac1{\ln2}.\]

So

\[\frac{a}{\ln2}=1,\]

and therefore

\[\boxed{a=\ln2}.\]

(ii) This is an exponential distribution with rate \(\ln2\). Hence

\[\boxed{E(X)=\frac1{\ln2}}.\]

(iii) For \(x\ge0\), the cumulative distribution function is

\[F(x)=1-e^{-x\ln2}.\]

The lower quartile \(Q_1\) satisfies \(F(Q_1)=\frac14\):

\[1-e^{-Q_1\ln2}=\frac14.\]

So

\[e^{-Q_1\ln2}=\frac34,\]

and

\[Q_1=\frac{\ln(4/3)}{\ln2}.\]

The upper quartile \(Q_3\) satisfies \(F(Q_3)=\frac34\):

\[1-e^{-Q_3\ln2}=\frac34.\]

So

\[e^{-Q_3\ln2}=\frac14,\]

and

\[Q_3=\frac{\ln4}{\ln2}=2.\]

Therefore the interquartile range is

\[Q_3-Q_1=\frac{\ln4-\ln(4/3)}{\ln2}=\frac{\ln3}{\ln2}.\]

Thus

\[\boxed{\mathrm{IQR}=\frac{\ln3}{\ln2}\approx1.58}.\]

(iv) Since \(Y=2^X\) and \(X\ge0\), we have \(Y\ge1\).

For \(y\ge1\),

\[G(y)=P(Y\le y)=P(2^X\le y).\]

This is

\[G(y)=P\left(X\le\frac{\ln y}{\ln2}\right).\]

So

\[G(y)=F\left(\frac{\ln y}{\ln2}\right)=1-e^{-\ln y}=1-\frac1y.\]

Differentiate to get the density:

\[g(y)=G'(y)=\frac1{y^2}.\]

Therefore

\[g(y)=\begin{cases}\dfrac1{y^2},&y\ge1,\\0,&y<1.\end{cases}\]