Thankyou so much...I've finally seen the light!
I made up a new, verified candidate list and factored in those eliminations (above) I could understand. Then, I proceeded as follows:
(1) a 3 must go somewhere in r9 in box 8, therefore 3 can be excluded from r9c9
(2) naked pairs 67 are in r7c8 and r9c9, therefore 6s and 7s can be excluded from the other squares in box 9
(3) a 7 must go somewhere in r8 in box 7, therefore 7s can be removed from r7 and r9 in box 7
(4) naked pairs 26 are in r4c2 and r7c2, therefore 6s can be excluded elsewhere in c2
That gives a placement of 6 in r3c3, which gives a placement of 2 in r2c3 and then the rest of the puzzle begins to unravel...
You guys are SO clever! Now, I don't feel quite so stupid.
Thanks again (Shazbot, QBasicMac, cecbevwr & em),
Paul.