(A) Let the number of days taken by a woman and a man be $x$ and $y$ respectively.
Therefore,the work done by a woman in $1$ day $= \frac{1}{x}$.
The work done by a man in $1$ day $= \frac{1}{y}$.
According to the problem:
$4(\frac{2}{x} + \frac{5}{y}) = 1 \Rightarrow \frac{2}{x} + \frac{5}{y} = \frac{1}{4}$ $(i)$
$3(\frac{3}{x} + \frac{6}{y}) = 1 \Rightarrow \frac{3}{x} + \frac{6}{y} = \frac{1}{3}$ $(ii)$
Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. Substituting these into the equations:
$2p + 5q = \frac{1}{4} \Rightarrow 8p + 20q = 1$ $(iii)$
$3p + 6q = \frac{1}{3} \Rightarrow 9p + 18q = 1$ $(iv)$
Solving by cross-multiplication:
$\frac{p}{(20)(-1) - (18)(-1)} = \frac{q}{(1)(9) - (1)(8)} = \frac{1}{(8)(18) - (20)(9)}$
$\frac{p}{-20 + 18} = \frac{q}{9 - 8} = \frac{1}{144 - 180}$
$\frac{p}{-2} = \frac{q}{1} = \frac{1}{-36}$
$p = \frac{-2}{-36} = \frac{1}{18}$ and $q = \frac{1}{-36} = -\frac{1}{36}$ (Wait,re-calculating: $p = \frac{1}{18}, q = \frac{1}{36}$).
Thus,$x = 18$ and $y = 36$.
Therefore,$1$ woman alone takes $18$ days and $1$ man alone takes $36$ days.