Kinetic Energy Alternative Expression 
Attempt: I've managed to show everything except for the last part. I need to show an alternative expression for $T$ as shown below. I've tried showing the relevant scalar product is the same as taking the scalar product of the velocity but it hasn't worked out, any help will be appreciated. 
 A: For the first part, we note that since $\vec B$ is constant 
$$m\vec r''=q\vec r'\times\vec B\implies m\vec r'=q\vec r\times\vec B \tag 1$$
Next, using $(1)$, we observe that
$$\begin{align}
\vec L\cdot \vec B&=(m\vec r \times \vec r')\cdot \vec B\\\\
&=-m\vec r'\cdot (\vec r\times \vec B)\\\\
&=-\frac{m^2}{q}(\vec r'\cdot \vec r') \tag 2
\end{align}$$
We also have 
$$\left|\vec r\times \vec B\right|^2=\frac{m^2}{q^2}(\vec r'\cdot \vec r') \tag 3$$
Therefore, using $(2)$ and $(3)$, we have 
$$\vec L\cdot \vec B+\frac12 q\left|\vec r\times \vec B\right|^2=-\frac12 \frac{m^2}{q}(\vec r'\cdot \vec r') \tag 4$$
Finally,
$$\begin{align}
\frac{d}{dt}\left(\vec r'\cdot \vec r'\right) &= 2\left(\vec r'\cdot \vec r''\right)\\\\
&=2\vec r'\cdot \frac{q}{m}(\vec r'\times \vec B)\\\\
&=0
\end{align}$$
so that $\vec L\cdot \vec B+\frac12 q\left|\vec r\times \vec B\right|^2$ is a constant of motion as was to be shown.

For the second part, we already showed that the Kinetic energy is a constant of motion since we showed that $$\frac{d}{dt}\left(\vec r\cdot \vec r\right)=0$$
Now we define a unit vector $\hat u=\frac{\vec r}{|\vec r|}$.  Since $\hat u$ is a unit vector, we observe that since $\hat u\cdot \hat u=1$, then 
$$\hat u'\cdot \hat u=\frac12 \frac{d}{dt}(\hat u\cdot \hat u)=0$$
and 
$$\frac12 \frac{d^2}{dt^2}(\hat u\cdot \hat u)=\hat u\cdot \hat u''+\hat u'\cdot \hat u'\implies \hat u\cdot \hat u''=-\hat u'\cdot \hat u' \tag 5$$
We also will make use of the identities 
$$r'=\frac{d}{dt}\left|\vec r\right|=\hat u\cdot \vec v \tag 6$$
and 
$$r \hat u' =\vec v-r'\hat u \tag 7$$
Then, using $(5)$, $(6)$, and $(7)$, we find 
$$\begin{align}
\hat u\cdot \left((\hat u\cdot \vec v)\vec v-r^2\hat u''\right)&=(\hat u\cdot \vec v)^2+r^2\hat u'\cdot \hat u'\\\\
&=(\hat u\cdot \vec v)^2+(\vec v-r'\hat u)\cdot (\vec v-r'\hat u)\\\\
&=(\hat u\cdot \vec v)^2+(\vec v\cdot \vec v-2r'\hat u\cdot \vec v+r'^2)\\\\
&=\vec v\cdot \vec v+(\hat u\cdot \vec v)^2-2(\hat u\cdot \vec v)^2+(\hat u\cdot \vec v)^2\\\\
&=\vec v\cdot \vec v
\end{align}$$
Therefore, we can write the kinetic energy $T=\frac12 m\vec v\cdot \vec v$ as
$$\bbox[5px,border:2px solid #C0A000]{T=\frac12 m\left(\hat u\cdot \left((\hat u\cdot \vec v)\vec v-r^2\hat u''\right)\right)}$$
as was to be shown!
