Answer
a) $\vec D=16\;\hat i−22\;\hat j$
b) See the figure below.
c) $27.2,\;306^\circ$
Work Step by Step
a) We need to find $\vec F$ in component forms whereas
$$\vec F=\vec A-4\vec B$$
Plugging the known;
$$\vec F=(4\;\hat i−2\;\hat j)-4(−3\;\hat i+5\;\hat j )=4\;\hat i+12\;\hat i−2\;\hat j-20\;\hat j$$
$$\boxed{\vec D=16\;\hat i−22\;\hat j}$$
b) See the figure below. We drew the 3 vectors there.
c) The magnitude of $\vec F$ is given by applying the Pythagorean theorem.
$$|\vec F|=\sqrt{F_x^2+F_y^2}=\sqrt{16^2+(-22)^2}=\color{red}{\bf 27.2}$$
and its direction is given by
$$\tan\alpha_F=\dfrac{F_y}{F_x}$$
Thus,
$$\alpha_F=\tan^{-1}\left[\dfrac{F_y}{F_x}\right]=\tan^{-1}\left[\dfrac{-22}{16}\right]=\bf -54^\circ$$
Thus,
$$\alpha_F=\color{red}{\bf 306^\circ}$$