(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2. 
$$
\begin{align*}
  (-8)^{4/3} &= (-8)^{1+1/3} \\
               &= -8\times(-8)^{1/3} \\
               &= -8\times (-1)^{1/3}\times 8^{1/3} \\
               &= -2\times 8\times (-1)^{1/3} \\
               &= -16\times (-1)^{1/3}.
\end{align*}
$$
So, which is the correct?

Both your solutions are correct. To see your first and second solution align note that one solution of $x = (-1)^{1/3}$ is $$x = -1$$
So one possible solution of your original problem is $$-16*(-1)^{1/3} = 16$$
This can be seen by observing that $$(-1)^3 = -1$$
In the complex plane $x= (-8)^{4/3}$ has multiple solutions of the form $-16r_i$ where $r_1,r_2,r_3$ are the complex cube roots of $-1$ which are 
\begin{align*}
r_1 &= \frac{1}{2} + \frac{\sqrt{3}}{2}i \\
r_2 &= \frac{1}{2} - \frac{\sqrt{3}}{2}i \\
r_3 &= -1
\end{align*}
