Write $|v(t)|$ as a piecewise function: $ |v(t)|= \cases{5-3t, & $0\le t\le 5/3$ \cr 3t-5, & $5/3\le t\le 3$ } $ Then, split the integral $\int_0^3 |v(t)|\,dt $ into two pieces.
$ \int_0^3 |v(t)|\,dt =\int_0^{5/3} |v(t)|\,dt +\int_{5/3}^3 |v(t)|\,dt =\int_0^{5/3} 5-3t\,dt +\int_{5/3}^3 3t-5\,dt, $ and evaluate.
Essentially, find the set over which $v$ is positive and the set over which $v$ negative; then integrate $|v|$ over these sets separately. On the set where $v$ is negative, you'd integrate $-v$ and over the set where $v$ is positive, you'd integrate $v$.
The difference between displacement and total distance should be clear: here the point travels to the left the first $5/3$ seconds then travels to the right from $t=5/3$ to $t=3$. The displacement is the difference between the final and initial position of the point. Since the point traveled left and then right, this will be less than the total distance traveled (which is the sum of the distance traveled left with the distance traveled right).
For example if you move along a line left $4$ units then right $5$ units, the displacement is $1$ and the total distance traveled is $9$.