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 $1\lt t\lt 2$.
Take $\vec{B} = \begin{pmatrix}0 \\ 0\\ k\end{pmatrix} k>0$
Now, $\vec{F} = q\vec{v} \times \vec{B} = q\begin{pmatrix}3 \\-1\\ 0\end{pmatrix}\times\begin{pmatrix}0 \\ 0\\ k\end{pmatrix} = -q\begin{pmatrix}k \\ 3k\\ 0\end{pmatrix}$
$$\hat{F} = -q\frac{k}{|q|k\sqrt{1+9}}\begin{pmatrix}1 \\ 3\\ 0\end{pmatrix} = \frac{-\operatorname{sgn}(q)}{\sqrt{10}}\begin{pmatrix}1 \\ 3\\ 0\end{pmatrix}$$
Try taking dot products $\vec{F}\cdot\vec{v}$ and $\vec{F}\cdot\vec{B}$
Best Answer
Let's call $\omega=\frac{qB_0}m$. Then your equations of motion are: $$\begin{align}\frac{dv_x}{dt}&=0\\\frac{dv_y}{dt}&=\omega v_z\\\frac{dv_z}{dt}&=-\omega v_y\end{align}$$ Note the sign on the last equation. The initial conditions for velocity can be written as $v_x(0)=v_0$, $v_y(0)=v_0$, and $v_z(0)=0$. The first equation gives $v_x(t)=v_0$, so $x(t)=v_0t$.
Take the derivative of the second equation, and use $\frac{d v_z}{dt}$ from the third to get $$\frac{d^2v_y}{dt^2}+\omega^2 v_y=0$$ The solution is $$v_y(t)=A\sin(\omega t)+B\cos(\omega t)$$ To find out $A$ and $B$ you need the initial conditions. First $$v_y(0)=B=v_0$$ Then: $$\frac{d v_y}{dt}(t)=\omega A\cos(\omega t)-v_0\omega\sin(\omega t)$$ At $t=0$ you have $$\frac{dv_y}{dt}(0)=\omega A=\omega v_z(0)=0$$ So you get $$v_y(t)=v_0\cos(\omega t)$$ and $$v_z(t)=-v_0\sin(\omega t)$$ Now integrate with respect to time, to get the positions. Using the initial conditions will shift the center for $z$