Answer
a) At $t=\rm 4 \;s,\; 8\; s$
b) See the graph below.
Work Step by Step
a) The author here asks to work with the graph, not with the kinematic equations.
So, the final position of the particle is given by the area under the $v_x-t$ curve if we know its initial position.
The author also told us that the particle starts from $x=0\;\rm m$ at $t=0\;\rm s$.
So that,
$$x_f=\overbrace{ x_0}^{{\color{red}=\;0}}+\text{Area under the curve}$$
$$x_f= \text{Area under the curve}$$
Now we can find the area under the curve by finding the area of one square. We can find that one square has 5 units height and 2 units wide. So, the area of one square is
$$A_{square}=5\;\rm m/s\cdot 2\;s=10\;\rm m$$
Now we need to find how many squares we need.
So,
$$N=\dfrac{35}{10}=3.5\;\rm square$$
Thus, at $t=4\;\rm s$, the particle reaches the position of 35 m.
Note that the particle will change its direction and move in the opposite direction. So, from $t=4\;\rm s$ to $t=6\;\rm s$ it moves away from the position of $x_f=35\;\rm m$ about 5 meters to the right (Note the half square area under the curve between $t=4\;\rm s$ and $t=6\;\rm s$). And then he moves the same distance of 5 m but in the opposite direction from $t=6\;\rm s$ to $t=8\;\rm s$ (Note the half square area under the curve between $t=6\;\rm s$ and $t=8\;\rm s$); which means that at $t=8\;\rm s$ he will be again at $x_f=35\;\rm m$.
From all the above, the particle will be at $x=35\;\rm m$ at
$$t= \color{red}{\bf4}\;{\rm s}, \;\;t=\color{red}{\bf8}\;\rm s$$
b)
In this part, we need to sketch the motion graph of the particle. We described its motion in previous part.
