Answer
$c$
Work Step by Step
$270^{\circ} < A< 360^{\circ}$
$\text{Dividing by }2\text{:}$
$135^{\circ} < \dfrac{A}{2} <180^{\circ}$
$\therefore \dfrac{A}{2} \text{ terminates in } QII$
$\cos{\dfrac{A}{2}} <0$
$\text{Using the half angle formula:}$
$$\cos{\dfrac{A}{2}} = -\sqrt{\dfrac{1+\cos{A}}{2}}$$
$\cos{\dfrac{A}{2}} = -\sqrt{\dfrac{1+\dfrac{2}{5}}{2} }= -\dfrac{\sqrt{70}}{10}$