$36 + 451 \div 11$

$(21.5 \% \text{ of } 999)^{1/3} + (42 \% \text{ of } 601)^{1/2} + ? = 28$ (Assuming the equation equals $28$ for a standard evaluation)

The sum of five consecutive even numbers of set-$A$ is $220$. What is the sum of a different set of five consecutive numbers whose second lowest number is $37$ less than double the lowest number of set-$A$?

The product of two natural numbers is $17$. Then the sum of the reciprocals of their squares is

$\frac{601}{49} \times \frac{399}{81} \div \frac{29}{201} = ?$

The digits indicated by $^*$ in $3422213^{**}$ so that this number is divisible by $99$ are