Answer: \(F(x)=0\) for \(x<2\), \(F(x)=\dfrac{(x-1)^2-1}{8}\) for \(2\le x\le4\), \(F(x)=1\) for \(x>4\); \(g(y)=\dfrac{1}{12y^{1/3}}\) for \(1\le y\le27\); median \(\boxed{5\sqrt5}\).

(i) For \(2\le x\le4\),

\[F(x)=\int_2^x \frac14(t-1)\,dt.\]

So

\[F(x)=\left[\frac18(t-1)^2\right]_2^x.\]

Therefore

\[F(x)=\frac{(x-1)^2-1}{8}.\]

Thus the distribution function is

\[F(x)=\begin{cases} 0, & x<2,\\ \dfrac{(x-1)^2-1}{8}, & 2\le x\le4,\\ 1, & x>4. \end{cases}\]

(ii) Since

\[Y=(X-1)^3,\]

we have

\[X=1+Y^{1/3}.\]

As \(2\le X\le4\), the range of \(Y\) is

\[1\le Y\le27.\]

Use the transformation formula:

\[g(y)=f(x)\left|\frac{dx}{dy}\right|.\]

Here

\[x=1+y^{1/3},\qquad \frac{dx}{dy}=\frac13y^{-2/3}.\]

Also

\[f(x)=\frac14(x-1)=\frac14y^{1/3}.\]

Therefore

\[g(y)=\frac14y^{1/3}\cdot\frac13y^{-2/3}=\frac{1}{12y^{1/3}}.\]

So

\[g(y)=\begin{cases} \dfrac{1}{12y^{1/3}}, & 1\le y\le27,\\ 0, & \text{otherwise}. \end{cases}\]

(iii) The distribution function of \(Y\) for \(1\le y\le27\) is

\[G(y)=P(Y\le y)=P(X\le 1+y^{1/3}).\]

Using the expression for \(F\),

\[G(y)=\frac{y^{2/3}-1}{8}.\]

Let the median be \(m\). Then \(G(m)=\frac12\), so

\[\frac{m^{2/3}-1}{8}=\frac12.\]

Hence

\[m^{2/3}=5.\]

Therefore

\[m=5^{3/2}=\boxed{5\sqrt5}.\]