Answer: \(F(x)=0\) for \(x<4\), \(F(x)=\dfrac1{40}(x^{3/2}-8x^{1/2}+8)\) for \(4\leq x\leq16\), and \(F(x)=1\) for \(x>16\). Also \(g(y)=\dfrac1{40}(3y^2-8)\) for \(2\leq y\leq4\), and \(0\) otherwise.

(i) For \(4\leq x\leq16\), integrate the density:

\[F(x)=\int \frac1{80}\left(3\sqrt{x}-\frac8{\sqrt{x}}\right)\,\mathrm dx.\]

This gives

\[F(x)=\frac1{80}\left(2x^{3/2}-16x^{1/2}\right)+C.\]

Use \(F(4)=0\):

\[0=\frac1{80}(2\cdot 4^{3/2}-16\cdot4^{1/2})+C=\frac1{80}(16-32)+C.\]

So \(C=\frac{16}{80}\). Therefore, for \(4\leq x\leq16\),

\[F(x)=\frac1{80}\left(2x^{3/2}-16x^{1/2}+16\right).\]

Equivalently,

\[F(x)=\frac1{40}\left(x^{3/2}-8x^{1/2}+8\right).\]

Hence

\[F(x)=\begin{cases} 0, & x<4,\\ \dfrac1{40}\left(x^{3/2}-8x^{1/2}+8\right), & 4\leq x\leq16,\\ 1, & x>16. \end{cases}\]

(ii) Since \(Y=\sqrt X\), the possible values of \(Y\) are

\[2\leq Y\leq4.\]

For \(2\leq y\leq4\),

\[G(y)=P(Y\leq y)=P(\sqrt X\leq y)=P(X\leq y^2)=F(y^2).\]

So

\[G(y)=\frac1{40}(y^3-8y+8).\]

Differentiate to get the density of \(Y\):

\[g(y)=G'(y)=\boxed{\frac1{40}(3y^2-8)},\qquad 2\leq y\leq4.\]

Outside this interval, \(g(y)=0\).