Lots of questions! We have $s(t)=t^3-6t^2+9t$. So if the velocity is denoted by $v(t)$, we have
$$v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3).$$
The particle is moving to the right when the velocity is positive, and to the left when the velocity is negative.
Looking at $3(t-1)(t-3)$, we note that it is positive when $t\gt 3$, also when $t\lt 1$. So in the time interval $(-\infty,1)$ and in the time interval $(3,\infty)$, to the degree this makes physical sense, we have motion to the right. In the time interval $(1,3)$ we have motion to the left.
The acceleration $a(t)$ is the derivative of velocity. So $a(t)=6t-12$.
There is some possible ambiguity (or trick) in the question about speeding up. The velocity is increasing when the acceleration is positive, that is, when $t\gt 2$. The velocity is decreasing when $t\lt 2$.
You should be able to do the rest of the parts. But "total distance travelled in the first $5$ seconds" is tricky, so we do some detail.
The net change in displacement is easy, it is $s(5)-s(0)$. But for total distance travelled, we need to take account of the fact that we are travelling to the right when $t$ is between $0$ and $1$, also when $t$ is between $3$ and $5$, while between $1$ and $3$ we are travelling to the left. So while $s(1)-s(0)$ and $s(5)-s(3)$ are positive, the number $s(3)-s(1)$ is negative.
Thus the total distance travelled in the first $5$ seconds is
$$|f(1)-f(0)|+|f(3)-f(1)|+|f(5)-f(3)|.$$
Maybe Don't Read: Velocity is not the same thing as speed. The speed at time $t$ is the absolute value of velocity, so it is $3|(t-1)(t-3)|$. We may want to know when speed is increasing. That's a different question than asking when velocity is increasing.
To find out you where speed is increasing, you can find out where the derivative of $(3(t-1)(t-3))^2$ is positive. This derivative is $9(t-1)(t-3)(2t-4)$. It is not hard to find out where this is positive: for $t\gt 3$ and for $1\lt t\lt 2$.
When you integrate the acceleration which is $a(t)=2t$ you obtain $v(t)=t^2 + C$ which is the constant of integration fixed by the initial conditions $v(0)=2$
Inserting the IC in the equation of the velocity you obtain that $C=2$ and therefore $v(t)=t^2 + 2$. Now find the antiderivative of this velocity and try to finish the problem.
Best Answer
For the motion of a particle we have: $$\boldsymbol a(t) = \frac{d\boldsymbol v(t)}{dt},~~~~~~~~\boldsymbol v(t) = \frac{d\boldsymbol s(t)}{dt}.$$ Now you should be able to find $s(t)$ and answer the questions.
Warning! Think about the last question: is the displacement $s(7)-s(4)$? Or is it the total amount of meters the particle travelled? Analogously: If I go from Washington to Buenos Aires, and then to Miami, was my displacement the distance between Washington and Miami? or was it the distance between Washington and Buenos Aires plus the distance between Buenos Aires and Miami?
Recall also that (this comes from Physics and from integrating $a=\frac{dv}{dt}$):
$$\boldsymbol s(t) = \boldsymbol s(0)+\boldsymbol v(0)t+\frac{1}{2}\boldsymbol a(t)t^2$$