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.
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!