Answer
a) $71.6^\circ$
b) $3,90^\circ$
c) $3,90^\circ$
Work Step by Step
a) To find the angle between $\vec E$ and $\vec F$, we need to find the angles that both made with the positive $x$-direction.
Thus,
$$\theta_E=\tan^{-1}\left(\dfrac{E_y}{E_x}\right)=\tan^{-1}\left(\dfrac{1}{1}\right)=\bf 45^\circ$$
And,
$$\theta_F=\tan^{-1}\left(\dfrac{F_y}{F_x}\right)=\tan^{-1}\left(\dfrac{1}{1}\right)= - 63.4^\circ$$
Noting that $\vec F$ is in the second quadrant.
So,
$$\theta_F=180-63.4=\bf 116.6^\circ$$
And hence, the angle between these two vectors is given by
$$\phi=\theta_F-\theta_E=116.6-45=\color{red}{\bf 71.6^\circ}$$
b) To find the direction and the magnitude of $\vec G=\vec E+\vec F$, we need to see the figure below.
From the geometry of the second figure below, we can see that the angle between $\vec E$ and $\vec F$ when we moved $F$ to the new position is given by $180^\circ -\phi$.
Now we can use the sin's law;
$$\dfrac{|\vec G|}{\sin (180^\circ -\phi)}=\dfrac{|\vec F|}{\sin45^\circ }$$
Plugging the known and solving for $|\vec G|$;
$$|\vec G|=\dfrac{ |\vec F|\sin (180^\circ -\phi)}{\sin45^\circ } =\dfrac{ \sqrt{F_x^2+F_y^2}\;\sin (180^\circ - 71.6^\circ)}{\sin45^\circ }$$
$$|\vec G|=\dfrac{\sqrt{(-1)^2+ 2^2}\;\sin (180^\circ - 71.6^\circ)}{\sin45^\circ }=\color{red}{\bf3}$$
And its direction relative to $+x$-direction is
$$\alpha_G=\color{red}{\bf90^\circ}$$
c)
Now we need to use the components method to find $\vec G$'s magnitude and direction.
$$\vec G=E_x\;\hat i+E_y\;\hat j+F_x\;\hat i+F_y\;\hat j$$
$$\vec G=1\;\hat i+1\;\hat j-1\;\hat i+2\;\hat j$$
$$\boxed{\vec G=0\;\hat i+3\;\hat j}$$
The magnitude of $\vec G$ is given by applying the Pythagorean theorem.
$$|\vec G|=\sqrt{G_x^2+G_y^2}=\sqrt{0^2+3^2}=\color{red}{\bf 3}$$
and its direction is given by
$$\tan\alpha_D=\dfrac{D_y}{D_x}$$
Thus,
$$\alpha_G=\tan^{-1}\left[\dfrac{G_y}{G_x}\right]=\tan^{-1}\left[\dfrac{3}{0}\right]=\bf 90^\circ$$
Thus,
$$\alpha_G=\color{red}{\bf 90^\circ}$$
