Hint: $\gcd(6,8)=2$.

It is more correct to write $\langle 6,8\rangle$ rather than $\langle \{6,8\}\rangle$, since the set $\{6,8\}$ is not an element of the group. And it is definitely more correct to write either one, rather than $<\{6,8\}>$.

@MishaLavrov I have definitely seen the notation $\langle S\rangle$ before, denoting the subgroup of $G$ generated by a subset $S\subset G$.

@KentaS That seems reasonable in the case that you're starting from a set $S$, but I still think it's not the best thing to do when you're writing out a set $S$ with curly braces.

Why did you not include $4k$ in your solution set?

@DerekHolt Since $4k\subset 2k$ , so there's no point in including it? Btw is the solution valid?...

OK, so why did you include $6k$, $8k$, and $14k$? (The answer is correct but not optimal.)

The correctness also depends on the grading rules. I would give $5$ out of $10$ points, mainly because it is not clear enough and could be written up much better. Moreover I suspect that the $\gcd$ has been used in the lecture, for dealing with such exercises.

Only senior professors are allowed to write "Clearly" in a mathematical proof :)

@Dietrich Burde Is the above solution valid?...

@DerekHolt I have edited it...does it fix the problem ?...

You wrote $8^n \in H$ which is true only for $n \ge 0$, and irrelevant.

@DerekHolt $8^n\in H$ is also true for $n\leq 0$ so that we could write $8k\in H$. Is the solution clear and valid?...

That's nonsense. $8^{-1} = 1/8$ is not in ${\mathbb Z}$ and $H$ is a subgroup of $({\mathbb Z},+)$.

@DerekHolt no, the group operation is $+$ here so $8^{-1}=-8$ and if the group operation is $×$ then $8^{-1}=1/8$ , but it is not so in case of the group $(\mathbb {Z},+)$ (since the identity element $e=0$ in this case)...

It's just a bad idea to mix additive and multiplicative notation for a group operation, all you'll do is to confuse anyone trying to understand you. If you choose $m+n$ to denote your group operation, then you should stick with $-n$ as the inverse operation.

Regarding correctness/validity of your solution, I'll echo some of the earlier comments: as it stands that $8^n$ makes your solution invalid. Correcting that would make it a valid solution. But it is like a plant that has been allowed to grow too much and become rangy, it needs a lot of trimming. So for all of those reasons, 5 points out of 10 seems about right.

@LeeMosher Thank you! But I still don't get why $8^n$ makes the solution invalid as $8^n=8n$ as the group operation is $+$ here and $8n\in H$ is indeed true..

@LeeMosher No, in group theory we define ,$a^n=a.a.a...a\space\text{n times}$ , where $.$ is the binary operation. Here , the binary operation is $+$ and hence, $8^n=8+8+8+...+8\space (\text {n times})=8n$. So, here $8^n$ essentially denotes $8n$.

@Franklin It is very difficult to help you if you keep contradicting people who are trying to correct your mistakes.

@DerekHolt how do u consider that a mistake ...u may consider it as a notational ambiguity or confusion as u may call it...(Also, I made it clear in the comment section about the notations )..At this point its clearer to me even , that "mistake" may be a wrong word used ...Thank you!...

Look I am trying to be helpful here, but you keep arguing. As Lee Mosher wrote, for the integer $8$, $8^n$ does not mean $8n$, and if you write $8^n$ meaning $8n$ in an exam, then you will certainly lose marks. In a structure such as ${\mathbb Z}$ which has both addition and multiplication already defined, you never use multiplicative notation within the additive group.