Answer
a) $\sqrt{2gh}$
b) $2.42\;\rm m/s$
Work Step by Step
a)
As we see in the figure below, we analyzed the weight of the ice box into its $x$ and $y$ components.
The net force exerted on the $ x$ direction is given by
$$\sum F_x=mg\sin\theta =ma_x$$
Thus, its acceleration is given by
$$ a_x= g\sin\theta $$
Noting that $\sin \theta=\dfrac{h}{\Delta x}$ where $\Delta x$ is the length of the ramp on which the box slides.
Hence,
$$ a_x= g\dfrac{h}{\Delta x}\tag 1 $$
The final speed of the box at the end of the ramp when it slides from rest is given by
$$v_{fx}^2=\overbrace{ v_{ix}^2 }^{0} +2a_x\Delta x$$
Plugging from (1);
$$v_{fx}^2=2 g\dfrac{h}{\Delta x}\Delta x$$
$$ v_{fx} =\color{red}{ \sqrt{2gh}} $$
b) As we see from the last formula above, the tilted angle of the ramp does not change the final velocity of the box, it is just the height.
So, the final velocity for both cases is
$$ v_{fx} = \sqrt{2gh}=\sqrt{2\cdot 9.8\cdot 0.30} $$
$$ v_{fx} =\color{red}{\bf 2.42}\;\rm m/s$$
