On 2.23 (a) you found a value of $n$ that made $20n$ a perfect square, so that after taking the square root $\sqrt{20n}$ was an integer. For part (b) you want to do something similar, but this time you want that after you take the square root the number you get is still a perfect square. Hint: What happens to the exponents in the prime factorization of a number when you take the square root of the number?

For 26, remember the factors of a number $n$ come in pairs that have a product equal to $n$, except when the number is a perfect square. For example $12 = 2^2\times3$ has factors $1$, $2$, $3$, $4$, $6$ and $12$, that can be paired as $1\times12 = 12$, $2\times 6 = 12$ and $3\times 4 = 12$. So, when you multiply all $6$ factors you get $$1\times2\times3\times4\times6\times12 = 12^3,$$ but if you do the same with a number like $10=2\times5$, that has only $4$ factors, you get only $$1\times2\times5\times 10 = 10^2$$ when you multiply together all of its factors.