6.1.2 problem 2

Let k be the median, therefore $P(X\ge k)\ge \frac{1}{2}$ and $P(X\le k)\ge \frac{1}{2}$, therefore $P(X\le k)\le \frac{1}{2}$.
Hence,

where F is the CDF of X.

Let c be the mode, $f$ be the PDF of X, since $f\left(x\right)$ is decreasing for all $x\ge 0$ ($f^{\text{'}}\left(x\right)=-{\lambda }^2{e}^{-\mathrm{\lambda x}}$) or is 0 otherwise, $f\left(x\right)$ has its maxima when $x=0$, hence, the mode is 0.