Explanation

(a)
If the order of reaction w.r.t. A is n and the order of reaction w.r.t. B is m, rate law become
Rate =$k\lbrack A\rbrack^{n}\lbrack B\rbrack^{m}$
From (1)
$5 \times 10^{- 5} = \lbrack 0.20\rbrack^{n}\lbrack 0.30\rbrack^{m}\ \ \ldots(i)$
From (2)
$5 \times 10^{- 5} = \lbrack 0.20\rbrack^{n}\lbrack 0.10\rbrack^{m}\ \ldots(ii)$
From (3)
$1 \times 10^{- 4} = \lbrack 0.40\rbrack^{n}\lbrack 0.05\rbrack^{m}\ \ \ldots(iii)$
$or\ \ \ \ 10 \times 10^{- 4} = \lbrack 0.40\rbrack^{n}\lbrack 0.05\rbrack^{m}$
from eqs.(i) and (ii)
$\frac{5 \times 10^{- 5}}{5 \times 10^{- 5}} = \left\lbrack \frac{0.20}{0.20} \right\rbrack^{n}\left\lbrack \frac{0.30}{0.10} \right\rbrack^{m}$
$1 = (3)^{m} \Rightarrow (3)^{0} = (3)^{m} \Rightarrow m = 0$
From eqs. (iii) and (iv)
$\frac{5 \times 10^{- 5}}{5 \times 10^{- 5}} = \left\lbrack \frac{0.20}{0.40} \right\rbrack^{n}\left\lbrack \frac{0.10}{0.05} \right\rbrack^{m}$
$\frac{1}{2} = \left( \frac{1}{2} \right)^{n} \times \left( \frac{0.10}{0.05} \right)^{0}$
$\Rightarrow \frac{1}{2} = \left( \frac{1}{2} \right)^{n} \Rightarrow \left( \frac{1}{2} \right)^{1} = \left( \frac{1}{2} \right)^{n} \Rightarrow n = 1$
Overall order of the reaction$= n + m = 1 + 0 = 1$