Answer
(a)$t_1>t_2$
(b) $t_1=1.73\,\mathrm{s}$
$t_2=0.72\,\mathrm{s}$
Work Step by Step
(a) $t_1>t_2$
The object begins accelerating from rest and so travels with a lower average speed in phase 1 than it does in phase 2. Hence it takes longer for it to cover the first $3.00\, \mathrm{m}$
(b) For phase 1, $x_1=3.00\,\mathrm{m}$, $x_{01}=0$, $v_{01}=0$ and $a=2.00\,\mathrm{m/s^2}$
Using the kinematic equation:
\begin{align*}
x_1&=x_{01}+v_{01}t_1+{1\over 2}at_1^2\\
3.00&=0+0+0.5\times 2.00 t_1^2\\
t_1^2&=3.00\\
t_1&=\sqrt{3.00}\\
t_1&=1.73\,\mathrm{s}
\end{align*}
For phase 2, $x_2=6.00\,\mathrm{m}$, $x_{02}=3.00\,\mathrm{m}$, $v_{02}=v_{01}+at_1=0+2.00(1.73)=3.46\,\mathrm{m/s}$ [i.e., the final velocity of phase 1 is the initial velocity of phase 2] and $a=2.00\,\mathrm{m/s^2}$.
\begin{align*}
x_2&=x_{02}+v_{02}t_2+{1\over 2}at_2^2\\
6.00&=3.00+3.46t_2+{1\over 2}(2.00)t_2^2\\
1.00t_2^2+3.46t_2-3.00&=0\\
\mathrm{Solving, we \,get}\\
t_2&=0.72\,\mathrm{s}
\end{align*}
Clearly $t_1>t_2$.
