Answer
a) See the figure below.
b) $\vec D=-(34.64\;{\rm m})\hat i+(15.36\;{\rm m})\hat j$
c) $37.89\;\rm m,156^\circ$
Work Step by Step
a) We drew the vector of the dog's motion, as you see below.
b) We used the graphical method to measure the net displacement of the dog, as you see in the second figure below.
Now we need to find the net displacement in component form.
$$\vec D =\vec A+\vec B+\vec C\\
\vec D =(A_x+B_x+C_x)\hat i+(A_y+B_y+C_y)\hat j\\\\
\vec D=(50\cos 45^\circ+70\cos180^\circ +20\cos 270^\circ )\hat i+(50\sin45^\circ+70\sin180^\circ +20\sin270^\circ)\hat j$$
Thus,
$$\boxed{\vec D=-(34.64\;{\rm m})\hat i+(15.36\;{\rm m})\hat j}$$
c)
The magnitude of $\vec D$ is given by applying the Pythagorean theorem.
$$|\vec D|=\sqrt{D_x^2+D_y^2}=\sqrt{(-34.64)^2+15.36^2}=\color{red}{\bf 37.89}\;\rm m$$
and its direction is given by
$$\tan\alpha_D=\dfrac{D_y}{D_x}$$
Thus,
$$\alpha_D=\tan^{-1}\left[\dfrac{D_y}{D_x}\right]=\tan^{-1}\left[\dfrac{15.36}{-34.64}\right]=\bf -23.9^\circ$$
And since $\vec D$ is in the second quadrant, so
$$\alpha_D=\color{red}{\bf 156^\circ}$$
