(a)

We have, $A_1, A_2, A_3 \ldots, A_m$ are non-empty subsets of $\{1,2,3, \ldots, 100\}$ $\left|A_1\right|,\left|A_2\right|, \ldots,\left|A_m\right|$ are distincts.

$A_1 \cap A_2 \cap A_3 \ldots \cap A_m=\phi$

$\therefore A_1 \cap A_2 \cup A_3 \ldots \cup A_m=\{1,2,3, \ldots, 100\}$

Let $\left|A_1\right|=1\left|A_2\right|=2 \ldots\left|A_m\right|=M$

$A_1, A_2, A_3 \ldots, A_m$ are disjoint set.

$\therefore\left|A_1\right|+\left|A_2\right| \ldots+\left|A_m\right|=100$

$1+2+3 \ldots+m=100$

$\frac{m(m+1)}{2}=100$

$m^2+m-200=0$

$=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-1 \pm \sqrt{1+4 \cdot 1 \cdot 200}}{2 \cdot 1}$

$=\frac{-1 \pm \sqrt{1+800}}{2}=\frac{-1+\sqrt{801}}{2}$

$=\frac{-1+28.30}{2}=\frac{27.30}{2}=16.65$

$m=\frac{1+28.30}{2}=\frac{29.30}{2}=14.65$

$\therefore m < 14$

$\therefore$ Maximum possible of $m$ is $13$ . ($14$th set will have same size as that of previous size)