Can I do this in Modulo arithmetic?

Question

The question says:

Find n for $(5n+6)8^{2n} \equiv 0 (mod 13)$ where $n$ is a normal number.

The book simply solved it by saying that since $8^{2n}$ and 13 are coprime(I dunno how though) then

$5n+6 = 0 (mod 13)$

And we find in the end $n = 4 (mod 13)$

But I didn't solve it this way because I didn't know how we can know if $8^{2n}$ is coprime with 13, I didn't it this way, I found it by

$(5n+6)8^{2n} \equiv 0 (mod 13)$

Which means $(5n+6)(-1)^{n} = 0(mod 13)$

We try for $n =2k+1$:

$(5(2k + 1) 6)(-1) = 0(mod13)$

Which means after calculating and all

$k = 8 (mod 13)$

We put k in n and we find

$n = 2(13 k + 8) + 1$ which is $n = 26k + 17$

Then for $n = 2k$(So -1 becomes 1 because it's an even number):

Same steps and we find $n = 26k + 4$

Now the thing I mostly ask about is, can I do this:

$n = 26k + 17$ which is $n = 13(2k) + 17$ and by defining $a = 2k$ then we write it $n = 13a + 17$ and since 13 is bigger than 17 we write it this way:

$n = 13a + 4$

And we do the same thing for $n = 26 k + 4$ which means $n = 13(2k) + 4$ and by putting $a = 2k$ then we can write it $n = 13a + 4$

And since we found the two values the same then we can say that n has a single value which is $n = 13a + 4$ which can be written

$n = 4 (mod 13)$ which is exactly as the book found it.

I have two questions:

1. How the book found that $8^{2n}$ and $13$ are coprime even though we don't know the exact value of n?

2. Can I solve it my way? I have doubts about it and I think it's wrong because I couldn't wrap my head around how I found $26k + 17 and 26k + 4$ in the first time(And they are clearly not equal) but then when I made them to $13a + 17$(Which is $13a + 4$) and $13a + 4$ they were clearly equal...how is that possible in math? I thought that if something is clearly not equal then no matter what you do and how you simplify it then you can't make them have the same form or make them equal.

Can anyone answer these two questions, please?