Answer: (a) \(k=3\).

(b) The median value is \(\dfrac{81}{16}\).

(c) The probability density function of \(Y\) is \(g(y)=\begin{cases}\dfrac{1}{8}y^2, & 0\leqslant y\lt 2,\\ \dfrac{2}{3}, & 2\leqslant y\leqslant 3,\\ 0, & \text{otherwise.}\end{cases}\)

(a) Since \(f\) is a probability density function, its total area must be \(1\).

So \(\displaystyle \int_0^4 \frac{1}{16}\sqrt{x}\,dx+\int_4^9 \frac{1}{k\sqrt{x}}\,dx=1\).

Now \(\displaystyle \int_0^4 \frac{1}{16}\sqrt{x}\,dx=\frac{1}{16}\left[\frac{2}{3}x^{3/2}\right]_0^4=\frac{1}{16}\cdot \frac{2}{3}\cdot 8=\frac{1}{3}\).

Also \(\displaystyle \int_4^9 \frac{1}{k\sqrt{x}}\,dx=\frac{1}{k}\left[2\sqrt{x}\right]_4^9=\frac{1}{k}(6-4)=\frac{2}{k}\).

Therefore \(\displaystyle \frac{1}{3}+\frac{2}{k}=1\), so \(\displaystyle \frac{2}{k}=\frac{2}{3}\), giving \(k=3\).

(b) Let the median be \(m\). Since \(P(X\leqslant 4)=\dfrac{1}{3}\), the median must lie in the interval \(4\leqslant m\leqslant 9\).

Using \(k=3\), we need \(P(X\leqslant m)=\dfrac{1}{2}\):

\(\displaystyle \int_0^4 \frac{1}{16}\sqrt{x}\,dx+\int_4^m \frac{1}{3\sqrt{x}}\,dx=\frac{1}{2}\).

The first integral is \(\dfrac{1}{3}\). For the second,

\(\displaystyle \int_4^m \frac{1}{3\sqrt{x}}\,dx=\frac{1}{3}\left[2\sqrt{x}\right]_4^m=\frac{2}{3}(\sqrt{m}-2)\).

Hence \(\displaystyle \frac{1}{3}+\frac{2}{3}(\sqrt{m}-2)=\frac{1}{2}\).

Simplifying, \(\displaystyle \frac{2}{3}\sqrt{m}-1=\frac{1}{2}\), so \(\displaystyle \frac{2}{3}\sqrt{m}=\frac{3}{2}\). Thus \(\displaystyle \sqrt{m}=\frac{9}{4}\), and therefore \(\displaystyle m=\frac{81}{16}\).

(c) Let the density of \(Y\) be \(g(y)\). Since \(Y=\sqrt{X}\), we have \(X=Y^2\). Also \(0\leqslant X\leqslant 9\), so \(0\leqslant Y\leqslant 3\).

For the transformation \(x=y^2\), \(\displaystyle \frac{dx}{dy}=2y\). Therefore, for \(y\geqslant 0\),

\(\displaystyle g(y)=f(y^2)\left|\frac{dx}{dy}\right|=2y f(y^2)\).

When \(0\leqslant y\lt 2\), we have \(0\leqslant y^2\lt 4\), so

\(\displaystyle g(y)=2y\cdot \frac{1}{16}\sqrt{y^2}=2y\cdot \frac{y}{16}=\frac{1}{8}y^2\).

When \(2\leqslant y\leqslant 3\), we have \(4\leqslant y^2\leqslant 9\), so, using \(k=3\),

\(\displaystyle g(y)=2y\cdot \frac{1}{3\sqrt{y^2}}=2y\cdot \frac{1}{3y}=\frac{2}{3}\).

Outside \([0,3]\), the density is \(0\). Hence

\(\displaystyle g(y)=\begin{cases}\dfrac{1}{8}y^2, & 0\leqslant y\lt 2,\\ \dfrac{2}{3}, & 2\leqslant y\leqslant 3,\\ 0, & \text{otherwise.}\end{cases}\)