For 3.21 it might help to look through some particular examples to get an idea of what's going on. On the example problems you've already found that factors most often come in pairs (for example, the factors of $12$ are $1$ and $12$, $2$ and $6$, and $3$ and $4$). When does it occur that a factor does not have a pair? 

For 3.26: Recall a number is a perfect square if it is an integer that has an integer square root, like $4$ or $169$. For a number to have an integer square root you need that all the exponents in its prime factorization are even (that way you can distribute the factors in two equal groups that yield the square root: $144 = 2^4 \times 3^2 = (2^2 \times 3)\times(2^2 \times 3)$). Something similar happens when a number is a perfect cube, only this time you will need that all the exponents are multiples of $3$. Now, for this problem you want $252\cdot x$ to be BOTH a perfect square and a perfect cube, so the exponents in its prime factorization need o be both multiples of $2$ AND multiples of $3$.

On 3.29 your first task is to find all proper divisors of $496$, that is, all numbers smaller than $496$ that divide $496$. Using the prime factorization of $496$ should help you find this divisors. Once you have them you just want to find the sum of all.

On 3. 30 you are doing the process of the Sieve of Eratosthenes for the number up to $1999$. You know that, if you do this process, at some point you will be left with only prime numbers on your list. For example, if you do it for the numbers up to $100$, $2$ will remove all even numbers, $3$ will remove some more, $5$ and $7$ will too, but $11$ will not remove any new numbers, so in this case it took $4$ prime numbers to remove all composite numbers up to $100$. (Hint: It might be worth noting what is the first number that each prime removes from the list.)