(a) Total number of cases obtained by taking multiplication of only two numbers out of $100 = {}^{100}{C_2}.$

Out of hundred $(1,\,\,2,\,………,\,100)$ given numbers,

there are the numbers $3,\,\,6,\,\,9,\,\,12,\,………,\,99,$ which are $33$ in number such that when any one of these is multiplied with any one of remaining $67$ numbers or any two of these $33$ are multiplied, then the resulting products is divisible by $3$.

Then the number of numbers which are the products of two of the given number are divisible by $3 = {}^{33}{C_1} \times {}^{67}{C_1} + {}^{33}{C_2}.$

Hence the required probability $ = \frac{{{}^{33}{C_1} \times {}^{67}{C_1} + {}^{33}{C_2}}}{{{}^{100}{C_2}}} = \frac{{2739}}{{4950}} = 0.55.$