Determining whether these groups are cyclic and listing the generators for the group 
Which of the following groups are cyclic:

*

*$\langle \mathbb{Z}, +\rangle$

*$\langle \mathbb{Q}, +\rangle$

*$\langle\mathbb{Q^+}, \cdot\rangle$

*$\langle \mathbb{6Z}, +\rangle$

*$\langle \{6^n\mid n \in \mathbb{Z}\}, \cdot\rangle$

*$\langle \{a + b \sqrt{2} \mid  a,b \in \mathbb{Z}\}, +\rangle$

I was typing out my answers to these when I realized I had no idea what I was doing, despite prior efforts. So any explanation of why these are cyclic or not cyclic is appreciated.
I know a group is cyclic if a generator can create the entire group, but I have no idea what I'm doing with regards to the actual generators.
Thanks for any help!
Reference: Fraleigh p. 56  Question 5.26 (sorry for typo) in A First Course in Abstract Algebra
 A: *

*Is $\mathbb{Z}$ a cyclic group under addition? The question you want to ask yourself is: is there an integer $x$ such that every integer is an integral multiple of $x$? (Note that under addition, the elements in the subgroup generated (not "created") by $x$ are precisely the integral multiples of $x$).
If so, the group is cyclic, generated by $x$. If not, then it is not cyclic. If it is cyclic, you can further ask What values of $x$ work? The answers are precisely the generators.


*Is $\mathbb{Q}$ cyclic under addition? Is there a rational number $q$ such that every rational number is an integral multiple of $q$?


*Is there a positive rational number $q$ such that every rational number is an integral power of $q$?


*Is there an integral multiple of $6$, $z$, such that every integral multiple of $6$ is an integral multiple of $z$?


*Is there an integral power of $6$, $w$, such that every integral power of $6$ is an integral power of $w$?


*Is there a number of the form $a+b\sqrt{2}$ such that every number of the form $x+y\sqrt{2}$ is an integral multiple of $a+b\sqrt{2}$?
When you think the answer is "yes", you should produce that distinguished element and show that every other element of your set is a multiple/power of that distinguished element. When you think the answer is "no", you should show that given any element, no matter what that element is, there is always some other element in the set that is not a multiple/power of the given element.
