For the first one, they are all integers, so first write $x \in \mathbb{Z}$. Then, notice that the set creates the series $1^2, 2^2, 3^2, ... 11^2$. Then, notice that this series are the perfect squares of some integer $y$ such that $1 <= y <= 11$. Therefore, we have to use something called an existential quantifier. Basically, these quantifiers let us make more complex statements in set-builder notation. For example, $\exists$ stands for the words "there exists." This means if I write:
$$(\exists y \in \mathbb{Z})(x=y^2)$$
I'm simply saying "there exists an integer $y$ such that $x=y^2$". Now, we want $1 <= y <= 11$, so we can add that to our statement:
$$(\exists y \in \mathbb{Z})(1<=y<=11 \wedge x=y^2)$$
This means "there exists an integer $y$ such that $1<=y<=11$ and $x=y^2$". That means $x$ is some perfect square from $1^2, 2^2, 3^2, ... 11^2$, which is exactly what we want. Using this statement, we can specify the set ${1, 4, 9, ..., 121}$ in set-builder notation:
$$\{x \in \mathbb{Z} | (\exists y \in \mathbb{Z})(1<=y<=11 \wedge x=y^2)\}$$
Now, for the second one, we need to find some set such that all of the elements of the set we are given are in that set. For example, if we were given ${1, 2, 3}$, then that set would be $\mathbb{Z}$ because $1$, $2$, and $3$ are all in $\mathbb{Z}$ since they are all integers. For this element, however, that's kind of tricky. We are given elements in the form of $(l, n)$ where $l$ is either $a$, $b$, or $c$ and $n$ is an integer. For $l$, we can figure out that $l \in {a, b, c}$ and for $n$, we can figure out that $c \in \mathbb{Z}$. However, we want to know what set $(l, n)$ is in. For this, we need to use a Cartesian product.
A Cartesian product lets us create sets that are ordered pairs of other sets. For example ${a, b, c} \times \mathbb{Z}$ is the set of all ordered pairs $(l, n)$ such that $l \in {a, b, c}$ and $n \in \mathbb{Z}$. This is exactly the set we are looking for such that all of our elements in our given set are in this set, so we've found the first part of the set-builder.
Now, we want to find the condition for the set-builder. If you look at all of the first coordinates, you will see that there is no condition because all of the elements of ${a, b, c}$ are covered in the first coordinate. However, in the second coordinate, you will see that all of the integers $n$ are such that $1 <= n <= 3$. Therefore, that is our condition. Thus, the set can be written in set-builder notation as follows:
$$\{(l, n) \in {a, b, c} \times \mathbb{Z} | 1 <= n <= 3\}$$
Thus, we are done with creating these set-builders for these sets. I hope this helped!
Best Answer
As written, the first. If you wanted to be less "wordy", you could write $x \equiv 0\ (\mathrm{mod}\ 5)$ instead of $x\ \mathrm{is \ divisible \ by \ }5$.
The second set you described would be empty, as the only $x$ satisfying $x = x/5$ is $x=0$, which is of course excluded by $x>0$. You could, however, rewrite that condition as $x/5 \in \mathbb{Z}$, which would describe the set you're after - again, just another way of writing it.