(a)

H3C−CH2−CH2−CH2−CH2−Br

1−Bromopentane

(b)

H3C−CH2−CH2− Br | CH −CH3

2−Bromopentane

(c)

H3C−CH2− Br | C | CH3 −CH3

2−Bromo−2−methyl‌butane

(i) These figures show that (a) contains the least steric hindrance so towards the SN2 reaction, 1-bromopentane will be the most reactive.
(ii) Figure b of 2-bromopentane has chiral carbon in it. So, it is optically active.
(iii) Towards the β-elimination, 2-bromo-2methylbutane will be the most reactive as it will form most stable alkene (on account of the highest number of α-hydrogens.)