Show that for all positive integers n, $\text{hcf}(6n + 8, 2n + 3) = 1$ To show that for all positive integers $n$, $\operatorname{hcf}(6n + 8, 2n + 3) = 1$
Can I just show that since n is a positive integer, so $6n + 8 > 4n + 5$.
When we apply long division
$6n + 8 = 1(4n + 5) + (2n +3)$
$4n + 5 = 1(2n + 3) + (2n +2)$
$2n + 3 = 1(2n + 2) + 1$
Therefore the highest common factor of $6n+8$ and $4n+5$ is $1.$

You can do what you've done; it's correct. Another way, but somewhat informal/intuitive...
$$6n + 8 = 2(3n+4)=6(n+1)+2=2(3(n+1)+1) => Even$$
$$2n + 3 = 2(n+1) + 1 => Odd$$
Now, it's not necessary that gcd(a,b) = 1 when a,b are odd and even respectively, but since they don't have a common factor after factorizing out, they're coprime. 
Your proof is rigorous, so it's fine really.
