The cycle will look like this (fingers crossed I can get the formatting anywhere near right)...
C + N2O -> CO + N2
....arrow.........arrow
...+1/2O2......+1/2O2
.....to............to
......CO + N2O
i.e. the two arrows pointing down represent:
C + 1/2O2 + N2O -> CO + N2O
and
CO + N2 +1/2O2 -> CO + N2O
You have two of the values, work out the third.
I teach delta T = final temp. - initial temp., i.e. temp. rises -> +ve delta T and temp. drops -> -ve delta T
q = m c delta T, i.e. temp rises -> +ve q and temp. drops -> -ve q (now I admit that you can't have a -ve q, but OCR (and I'd bet other exam boards) will overlook this).
Then delta H = - q / n, i.e. temp. rises -> -ve delta H and temp. drops -> +ve delta H (the minus sign for exothermic reactions no longer appears from nowhere).
For your two reactions, both involve temp. rises and therefore both should have been exothermic, i.e. -ve delta H.