$\therefore [{H^ + }] = \frac{{20 \times 2}}{{800}} = \frac{1}{{20}} \Rightarrow pH =  - log\left( {\frac{1}{{20}}} \right)$

$\therefore \,pH = 1.3\,$ so $(a)$ is correct

$(b)\,\log \,\left( {\frac{{{K_{{w_2}}}}}{{{K_{{w_1}}}}}} \right) = \frac{{\Delta H}}{{2.303\,R}}\left[ {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right]$

so ionic product of water is temp. dependent hence $(b)$ is correct.

$(c)\,{K_a} = {10^{ - 5}},\,pH = 5 \Rightarrow [{H^ + }] = {10^{ - 5}}$

${K_a} = \frac{{c{\alpha ^2}}}{{(1 - \alpha )}} \Rightarrow {K_a} = \frac{{{{[H]}^ + }.\alpha }}{{(1 - \alpha )}}$

$\therefore \,{10^{ - 5}} = \frac{{{{10}^{ - 5}}.\alpha }}{{(1 - \alpha )}} \Rightarrow 1 - \alpha  = \alpha  = \frac{1}{2} = 50\% $

so $(c)$ is correct

$(d)$ Le-chatelier's principle is applicable to cpmmon-Ion effect so option $(d)$ is wrong