How to reparametrize with respect to arc length? 
In part (a), I calculated that s(t) = $\int_0^t \! (\frac{t^2}{2}  + 2t) \, \mathrm{d}t$ = $t^2/2 + 2t$
I'm unsure how to solve part (b). My attempt is:
$t^2/2 + 2t$ = s $\iff$ t = -2 $\pm$ $\sqrt{2s+4}$
r(s) = r(t(s)) = {2(-2 $\pm$ $\sqrt{2s+4}$), $\frac{4}{3}$(-2 $\pm$ $\sqrt{2s+4}$)^(3/2), 0.5$\sqrt{2s+4})^2$}
 A: Notice that even after the edit your solution for $(a)$ is only almost correct. First, you are asked to find the length of the entire curve, second the integrand is incorrect (the final result coincide with what I have found, so this might be just a typo now that you have improved your answer.) I should mention that it is advisable to use different names for variables of integration and the integration's extremes.
For part $(b)$, you need to pick the solution which is positive (notice that $\textbf r$ is defined on $[0,4]$, so can only evaluate it when $s$ is positive).
The following solution was written to address the question before the edit.

The formula for the length of the a parametric curve $C$ is $$\int_a^b|\textbf{r}'(t)|\,dt, \tag 1$$
where $[a,b]$ is the interval of definition of the parametrization $\textbf r$. (in our case $a = 0$ and $b = 4$).
The first thing to do is to compute $\textbf{r}'(t)$. Recall that this can be achieved by differentiating each component: $$\textbf r'(t) =  \left[
    \begin{array}{cc|c}
      2\\
      2\sqrt t\\
      t
    \end{array}
\right].$$
Next let's compute $|\textbf{r}'|$: $$|\textbf{r}'(t)| = \sqrt{2^2 + (2\sqrt t)^2 + t^2} = \sqrt{4 + 4t + t^2} = t + 2.$$ Finally, we can plug this into $(1)$ to find the length of $C$: $$L = \int_0^4 2 + t\, dt = 16.$$
To solve part $(b)$ we'll need the formula for arc length, which is $$s(t) = \int_0^t|\textbf{r}'(\tau)|\,d\tau = \int_0^t 2 + \tau\, d\tau = \frac{1}{2}t^2 + 2t.$$ We are interested in finding $t$ in terms of $s$, hence we need to solve the second order equation $$\frac{1}{2}t^2 + 2t - s = 0.$$ This has two solutions, we discard the negative one and keep $$t(s) = \sqrt{2s + 4} - 2.$$ Then let $$\textbf r(s) =  \left[
    \begin{array}{cc|c}
      2\sqrt{2s + 4} - 2\\
      \frac{4}{3}(\sqrt{2s + 4} - 2)^{\frac 32}\\
      \frac 12(\sqrt{2s + 4} - 2)^2
    \end{array}
\right].$$
This is a parametric representation of the curve $C$ with respect to arc length.
