Answer
a) $ \vec E=10\;\hat i+2\;\hat j $
b) See the figure below.
c) $10.2,\;11.3^\circ$
Work Step by Step
a) We need to find $\vec E$ in component forms whereas
$$\vec E=4\vec A+2\vec B$$
Plugging the known;
$$\vec E=4(4\;\hat i−2\;\hat j)+2(−3\;\hat i+5\;\hat j )=16\;\hat i-6\;\hat i−8\;\hat j+10\;\hat j$$
$$\boxed{\vec E=10\;\hat i+2\;\hat j}$$
b) See the figure below. We drew the 3 vectors there.
c) The magnitude of $\vec E$ is given by applying the Pythagorean theorem.
$$|\vec E|=\sqrt{E_x^2+E_y^2}=\sqrt{10^2+2^2}=\color{red}{\bf 10.2}$$
and its direction is given by
$$\tan\alpha_E=\dfrac{E_y}{E_x}$$
Thus,
$$\alpha_E=\tan^{-1}\left[\dfrac{E_y}{E_x}\right]=\tan^{-1}\left[\dfrac{2}{10}\right]=\bf 11.3^\circ$$
Thus,
$$\alpha_E=\color{red}{\bf 11.3^\circ}$$
