1).
We remind that : $FDR=\mathbb{E}[\frac{FP}{TP+FP}1_{TP+FP\geqslant 1}]$ and $FWER=\mathbb{P}(FP>0)=\mathbb{E}[1_{FP>0}]$.
-If $TP+FP\geqslant 1$, then $\frac{FP}{TP+FP}\leqslant 1$ is always satisfied. In particular $\frac{FP}{TP+FP}=0$ when $FP=0$.
-If $TP+FP=0$, then $FP=0$ (because $TP\geqslant 0$ and $FP\geqslant 0$).
In both cases we can conclude that $\frac{FP}{TP+FP}\leqslant 1_{FP>0}$, and by taking $\mathbb{E}$, which is a non decreasing fonction for positive functions, we proved that $FDP \leqslant FWER$.
2).
The Bonferroni procedure ensures that $FDP \leqslant \alpha$, the calculus appears page 169 of the book.
3).
The number of rejected hypothesis is given by $Card(\hat{R})$.
In the Bonferroni procedure, $Card(\hat{R}_{Bonf})=Card(\{k \in I, \hat{p}(k)\leqslant \frac{\alpha}{m}\})$. Whereas in the Benjamini-Hochberg procedure, $Card(\hat{R}_{BH})=Card(\{k \in I, \hat{p}(k)\leqslant \frac{\alpha k}{m}\})$.
Knowing that $k\geqslant 1$, it is clear that $\hat{R}_{Bonf}\subset \hat{R}_{BH}$ and as a consequence we have that $Card(\hat{R}_{Bonf})\leqslant Card(\hat{R}_{BH})$.
4).
The number of rejected hypothesis for the Benjamini-Yekutieli procedure is $Card(\hat{R}_{BY})=Card(\{k \in I, \hat{p}(k)\leqslant \frac{\alpha k}{m H_m}\})$.
Noticing that $\hat{R}_{Bonf}\subset \hat{R}_{BY}$ is equivalent to ask that $\frac{\hat{k}}{H_m}\geqslant 1$. We conclude that the Benjamini-Yekutieli procedure rejects more hypothesis than the Bonferroni procedure only when $\hat{k}\geqslant H_m$.
Solution proposed by Tabouy T.