(a) Ethylene chloride and ethylidene chloride on reaction with with ale. KOH give elimination reaction and ethyne as the product.

$Cl—C{H}_{2}C{H}_2—Cl\underset{KOH\left(excess\right)}{\overset{Alc.}{\to }}CH\equiv CH$
$Ethylene chloride Ethyne$
$$
$C{H}_3—CHC{l}_2\underset{\left(Excess\right)}{\overset{Alc. KOH}{\to }}CH\equiv CH$
$Ethylidene chloride$
$$

(b) Both these compounds form different products on treatment with aq. NaoH 

$Cl—C{H}_2—C{H}_{2}Cl\underset{NaOH}{\overset{Aq.}{\to }}HO—C{H}_2—C{H}_2—OH$
$Ethylene glycol$
$C{H}_3—CHC{l}_2\underset{NaOH}{\overset{Aq.}{\to }}C{H}_3—CH{\left(OH\right)}_2\underset{-H_{2}O}{\to } C{H}_{3}CHO$
$Ethanal$

(c) Both these compounds form same products on reduction.

$Cl—C{H}_2—C{H}_2—Cl\stackrel{Reduction}{\to }{H}_{3}C—C{H}_3—2HCl$
$Ethane$
$C{H}_3—CHC{l}_2\stackrel{Reduction}{\to }{H}_{3}C—C{H}_3+ 2HCl$
$Ethanal$

(d) Both these compounds are optically inactive.