4. Решение:
\[ \angle MCF + \angle FCN = 180^{\circ} \]
\[ x + \frac{x}{3.5} = 180^{\circ} \]
\[ \frac{3.5x + x}{3.5} = 180^{\circ} \]
\[ \frac{4.5x}{3.5} = 180^{\circ} \]
\[ 4.5x = 180^{\circ} \times 3.5 \]
\[ 4.5x = 630^{\circ} \]
\[ x = \frac{630^{\circ}}{4.5} \]
\[ x = 140^{\circ} \]
\[ \angle MCF = x = 140^{\circ} \]
\[ \angle FCN = \frac{x}{3.5} = \frac{140^{\circ}}{3.5} = 40^{\circ} \]
Ответ: ∠MCF = 140°, ∠FCN = 40°.