Answer: \(g(y)=\dfrac1{10}(3y-1)\) for \(1\leq y\leq3\), and \(\boxed{E(Y)=\dfrac{11}{5}}\).

(i) First find the distribution function of \(X\) for \(1\leq x\leq9\):

\[F(x)=\int_1^x\frac1{20}\left(3-\frac1{\sqrt{u}}\right)\,\mathrm du.\]

Integrating gives

\[F(x)=\frac1{20}\left[3u-2\sqrt{u}\right]_1^x.\]

So

\[F(x)=\frac1{20}(3x-2\sqrt{x}-1).\]

Now \(Y=\sqrt X\). Since \(1\leq X\leq9\), the possible values of \(Y\) are

\[1\leq Y\leq3.\]

For \(1\leq y\leq3\),

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

Therefore

\[G(y)=\frac1{20}(3y^2-2y-1).\]

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

\[g(y)=G'(y)=\frac1{20}(6y-2)=\frac1{10}(3y-1).\]

Thus

\[g(y)=\begin{cases} \dfrac1{10}(3y-1), & 1\leq y\leq3,\\ 0, & \text{otherwise.} \end{cases}\]

(ii) The mean of \(Y\) is

\[E(Y)=\int_1^3 y g(y)\,\mathrm dy.\]

So

\[E(Y)=\frac1{10}\int_1^3 y(3y-1)\,\mathrm dy.\]

That is

\[E(Y)=\frac1{10}\int_1^3(3y^2-y)\,\mathrm dy.\]

Hence

\[E(Y)=\frac1{10}\left[y^3-\frac12y^2\right]_1^3.\]

Evaluate:

\[E(Y)=\frac1{10}\left((27-4.5)-(1-0.5)\right)=\frac1{10}(22)=\boxed{\frac{11}{5}}.\]