(a) A density integrates to 1. Hence \[1=k\int_0^2(2x^2-x^3)\,dx=k\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2=k\left(\frac{16}{3}-4\right)=\frac{4k}{3},\] giving \(k=3/4\).

(b)(i) By the definition of the median for this continuous distribution, \(\mathrm P(X\le m)=0.5\).

(b)(ii) First find the mean: \[\mathrm E(X)=\frac34\int_0^2(2x^3-x^4)\,dx=\frac34\left[\frac{x^4}{2}-\frac{x^5}{5}\right]_0^2=\frac65.\] Then \[\mathrm P(X\le6/5)=\frac34\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^{6/5}=\frac{297}{625}=0.4752.\] This is less than 0.5, so the mean lies below the median. Therefore \[\mathrm P(\mathrm E(X)\le X\le m)=0.5-0.4752=0.0248=\frac{31}{1250}.\]