Cu+HNO2-Cu(NO3)2+NO solved by oxidation no. Method

Answer 1

You made a typo. You meant HNO3 and you omitted H2O as one of the products.
Cu + HNO3-Cu(NO3)2 + NO + H2O

Cu on left has oxidation number 0; on right is +2. N on left is +5 and on right is +2 (in NO) and for the moment ignore the NO3 ion in Cu(NO3)2.

So multiply the Cu change by 3 and the N change by 2 to make 6 electrons lost and gained; that gives you
3Cu + 2HNO3 --> 3Cu(NO3)2 + 2NO + H2O
That balances the oxidation-reduction part. Now you notice there are 6 NO3^- in which the N did NOT change so add 6 more to the HNO3 on the left to handle that. Now the equation is
3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO + H2O and then balance the H by making H2O 4. Done.
3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO + 4H2O