Two numbers are chosen from {1,2,3,4,5,6,7,8}one after another without replacement. Then the probability that

n(s) = 8 / 2 = 28 
a) Smaller one less than 3$\Rightarrow$ smaller are can be 2 or 1 c the second can be selected in 6 or 7 ways respectively probability  $=\frac{6+7}{28}=13/28$
b) bigger value of the two is more than 5$\Rightarrow$ bigger one can be 6,7,8 respectively the second number can be in 5,6,7 ways c) respectively probability  $=\frac{5+6+7}{28}=\frac{18}{28}$
product is even $\Rightarrow$at least one of them is even probability $=1-\frac{4{\mathrm{c}}_2}{8{\mathrm{c}}_2}=1-\frac{6}{28}=\frac{11}{14}$