Any quick method of doing this one - or do I have to write down 1/7,2/7.....6/7 and 1/5....4/5 and then subtract every one of them?

Stacy / Ron : A shortcut or underlying concept for such sums would be appreciated

the idea here is that you want to perform a mutual comparison of ALL those fractions at the same time.

here's the golden rule:
to compare more than two fractions at a time, make the common denominator of all the fractions in question.

here, since you literally want to compare ALL these fractions at once, there's no doubt that you want to make the common denominator of all the fractions in the problem: 35.

the fifths become 7/35, 14/35, 21/35, 28/35.
the sevenths become 5/35, 10/35, 15/35, 20/35, 25/35, 30/35.
the smallest distance between any pair of these is 1/35 (between 20/35 and 21/35, or between 14/35 and 15/35).
so answer = b.

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you could also predict the answer to this problem: you know that it's going to have a 35 in the denominator - you can't subtract fifths and sevenths and get a fraction whose denominator is 70, 12, or 7 - and you should be highly suspicious of (c) because that's just the difference between 1/5 and 1/7. therefore, you'd go with (b).
of course, the systematic method proffered above is preferable.