Answer
a) $ \vec C=1\;\hat i+3\;\hat j $
b) See the figures below.
c) $3.16,\; 71.6^\circ $
Work Step by Step
a) We need to find $\vec C$ in component forms whereas
$$\vec C=\vec A+\vec B$$
Plugging the known;
$$\vec C= \left(4\;\hat i-2\;\hat j\right)+ \left(-3\;\hat i+5\;\hat j\right) =4\;\hat i-3\;\hat i-2\;\hat j+5\;\hat j$$
$$\boxed{\vec C=1\;\hat i+3\;\hat j}$$
b) See the figures below. We drew the 3 vectors there.
c) The magnitude of $\vec C$ is given by applying the Pythagorean theorem.
$$|\vec C|=\sqrt{C_x^2+C_y^2}=\sqrt{1^2+3^2}$$
$$|\vec C|=\color{red}{\bf 3.16}$$
and its direction is given by
$$\tan\alpha_C=\dfrac{C_y}{C_x}$$
Thus,
$$\alpha_C=\tan^{-1}\left[\dfrac{C_y}{C_x}\right]=\tan^{-1}\left[\dfrac{3}{1}\right]$$
$$\alpha_C=\color{red}{\bf 71.6^\circ}$$