This question isn't clear. I'm assuming that if there are errors in A and B (which we'll call εA and εB) then we get a corresponding error εC in C.
(1) So we can write C+εC=(A+εA)(B+εB)=AB+AεB+BεA+εAεB.
The last term can be ignored because for small errors it will be negligible.
On the RHS we can replace AB by C, and we obtain an expression for εC:
εC=AεB+BεA.
(2) We can write C+εC=(A+εA)/(B+εB).
1/(B+εB)=(B+εB)-1=B-1(1+εB/B)-1=(1-εB/B)/B approx.
So:
C+εC=(A+εA)/(B+εB)=[(A+εA)/B](1-εB/B)=(A-AεB/B+εA-εAεB/B)/B. We can ignore εAεB/B as negligible:
C+εC=(A/B)+(εA-AεB/B)/B=C+(εA-AεB/B)/B, εC=(εA-AεB/B)/B or (BεA-AεB)/B2. Note that errors are usually prefixed by ± so the error in C will also be ±.
If errors are given in fractions εA=ApA, εB=BpB. For percentage errors εA=ApA/100, εB=BpB/100, where pA and pB are fractions or percentages as appropriate.
I hope that this is in accordance with the intention of the question.