Answer
(a) $v_1>{1\over 2}v_2$
(b) $v_1=9.22\,\mathrm{m/s}$ and $v_2=13.0\,\mathrm{m/s}$
Work Step by Step
(a) $v_1>{1\over 2}v_2$
This basically means that $v_2$ (the speed at the end of phase 2) is less than twice $v_1$ (the speed at the end of phase 1).
I.e., the final speed $v$ does not double if the displacment ($\Delta x$) is doubled. This can be seen with the help of the kinematic equation,
$$v^2=v_0^2+2a\Delta x$$
If we set $v_0=0$ then $v^2=2a\Delta x$, or $v=\sqrt{2a\Delta x}$.
Now, if $x\rightarrow 2x$, then $v\rightarrow \sqrt{2}v$, which is about 1.4 times v--less than twice of v.
Hence, $v_2<2v_1$ or $v_1>{1\over 2}v_2$.
(b) For phase 1, $v_{01}=0$, $a=0.850\,\mathrm{m/s^2}$, $x_{01}=0$ and $x_1=50\,\mathrm{m}$.
Using the kinematic equation
\begin{align*}
v_1^2&=v_{01}^2+2a(x_1-x_{01})\\
&=0+2(0.850)(50-0)\\
&=85.0\\
v_1&=\sqrt{85.0}\\
&=9.22\,\mathrm{m/s}
\end{align*}
For phase 2, $v_{02}=v_1=9.22\mathrm{m/s}$, $a=0.850\,\mathrm{m/s^2}$, $x_{02}=50\,\mathrm{m}$, $x_2=100\,\mathrm{m}$. And so,
\begin{align*}
v_2^2&=v_{02}^2+2a(x_2-x_{02})\\
&=9.22^2+2(0.850)(100-50)\\
&=85.0+85.0\\
&=170\\
v_2&=\sqrt{170}\\
&=13.0\,\mathrm{m/s}
\end{align*}
And so $v_1>{1\over 2}v_2$ since $9.22>{1\over 2}(13.5)$.
