If k = 2n - 1, where n is an integer, what is the remainder

This one can be done very easily with number picking. As you correctly noted \(k = 2n - 1\) means that k is an odd number (basically \(k = 2n - 1\) is a formula of an odd number).

Now let's try several odd numbers:
k=1 --> k^2=1 ---> remainder upon division of 1 by 8 is 1;
k=3 --> k^2=3 ---> remainder upon division of 9 by 8 is 1;
k=5 --> k^2=25 ---> remainder upon division of 25 by 8 is 1;

At this point we can safely assume that this will continue for all odd numbers.

But if you want algebraic approach, here you go:
\(k = 2n - 1\) --> \(k^2=(2n-1)^2=4n^2-4n+1=4n(n-1)+1\) --> what is a remainder when \(4n(n-1)+1\) is divided by 8:

Now either \(n\) or \(n-1\) will be even so in any case \(4n(n-1)=4*odd*even=multiple \ of \ 8\), so \(4n(n-1)\) is divisible by 8, so \(4n(n-1)+1\) divided by 8 gives remainder of 1.

Answer: A.

Hope it's clear.