The first reaction

`Cu(s) + 4HNO_3 (aq) -> Cu(NO_3)_2 (aq) + NO_2 (g) + H_2O(l)`

is an oxidation-reduction reaction. The oxidation state of copper is 0 on the reactant side and +2 on the product side. The loss of 2 electrons indicates that copper has been oxidized. Similarly, nitrogen has an oxidation state of +5 on the reactant side and +4 on the product side (in NO2 gas). The gain of an electron indicates that it has been reduced. 

The second reaction

`Cu(NO_3)_2 (aq) + 2 NaOH (aq) -> Cu(OH)_2 (s) + 2NaNO_3 (aq)`

is a double displacement reaction, since the reacting species exchange their cations and anions. On the reactant side, copper is bonded with nitrate ionz and sodium is bonded with a hydroxide ion. However, on the product side, copper is bonded with hydroxide ions and sodium is bonded with a nitrate ion. Thus, a double displacement has taken place.

Hope this helps.