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 $1\lt t\lt 2$.
Your part $(ii)$ is wrong. For instance, assuming $t \geq 0$, $x'(t) < 0$ on the interval $(\frac{1}{2},+\infty)$, not $(0, \frac{1}{2})$. Have a look at this part again, then you can fix part $(iii)$.
Best Answer
You've got the right idea, but not the right execution. Let's call the direction of positive $x$ "up" and the opposed direction down.
It is moving upward--has a positive velocity--on $(0,1)$ and $(3,4)$. It is moving downward--has a negative velocity--on $(1,3).$ Put another way, the graph's height is increasing on $(0,1)$ as we move to the right--likewise in $(3,4)$--and decreasing on $(1,3)$ as we move to the right.
It is accelerating upward--has a positive acceleration--on $(2,4)$. It is accelerating downward--has a negative acceleration--on $(0,2)$. Put another way, the graph's slope is increasing on $(2,4)$ as we move to the right, and the graph's slope is decreasing on $(0,2)$ as we move to the right.
Now check where the signs match.