(i) $\ce{NO}$: +2 ✔; $\ce{NO2}$: +4 ✔

(ii) $K_c = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$ ✔

(iii) exothermic reaction «$\Delta H^\ominus$ negative», so increasing $T$ shifts the position of equilibrium towards the reactants «the endothermic direction» and $K_c$ decreases ✔

(iv) Position of equilibrium: decreasing the volume increases the pressure/concentration; the system shifts towards the side with fewer gas molecules, i.e. towards the products «2 mol $\ce{NO2}$ vs 3 mol of gas on the reactant side», so the position of equilibrium moves to the right ✔

Value of $K_c$: unchanged, because $K_c$ depends only on temperature, which has not changed ✔

(i) [1] both oxidation states correct for the mark. (ii) [1] correct expression, NO squared and O2 not squared. (iii) [1] single marking point: K decreases, with a correct reason (exothermic/Le Chatelier); both the direction and the reason are needed for the mark. (iv) [2] M1: correct position of equilibrium (shifts towards products) with a reason based on the number of gas molecules; M2: correct statement that Kc is unchanged, because only temperature changes Kc.