Answer
a) $2.56\;\rm m/s$
b) $33\;\rm cm$
Work Step by Step
We choose the system to be the two packages plus the Earth, the left one and the right one. We can assume that this system is isolated since the left one slide without friction from rest.
We need to find the speed of the left package just before it hits the other one.
We can use the conservation of energy for this package since we can consider it plus the Earth as an isolated system.
$$E_{top}=E_{ground}$$
$$U_{g, top}+\overbrace{K_{top}}^{0}=\overbrace{U_{g,ground}}^{0}+K_{ground}$$
where its initial velocity is zero, so its initial kinetic energy is zero as well. And we chose the bottom of the path to be the zero height of $y=0$, so the final gravitational potential energy is zero.
$$U_{g, top}=K_{ground}$$
$$\color{red}{\bf\not} mgh=\frac{1}{2}\color{red}{\bf\not} mv^2$$
Solving for $v$;
$$v=\sqrt{2 gh}=\bf 7.67\;\rm m/s\tag 1$$
a) Now we can consider that the momentum is conserved during the collision between the two packages.
$$p_{ix}=p_{fx}$$
$$\color{red}{\bf\not} mv=(m+2m)v_f=3\color{red}{\bf\not} mv_f$$
where the two packages stuck together after the collision.
$$ v=3v_f$$
$$v_f=\dfrac{v}{3}$$
Plugging from (1);
$$v_f=\dfrac{ \sqrt{2 gh}}{3}$$
Plugging the known;
$$v_f=\dfrac{ \sqrt{2 \times 9.8\times 3}}{3}=\color{red}{\bf 2.56}\;\rm m/s$$
b) For a totally elastic collision, the momentum is also conserved, but the formula will be as follows.
$$ \color{red}{\bf\not} mv= \color{red}{\bf\not} mv_1+2 \color{red}{\bf\not} mv_2$$
where $v_1$ is the final speed for $m$, and $v_2$ is the final speed for $2m$.
$$ v= v_1+2 v_2 $$
Hence,
$$v_2=\dfrac{ v- v_1}{2}\tag 2$$
The energy is also conserved, so
$$\color{red}{\bf\not}\frac{1}{2}\color{red}{\bf\not}mv^2=\color{red}{\bf\not}\frac{1}{2}\color{red}{\bf\not}mv_1^2+\color{red}{\bf\not}\frac{1}{2}(2\color{red}{\bf\not}m)v_2^2$$
Thus,
$$v^2= v_1^2+ 2v_2^2$$
Plugging from (2);
$$v^2= v_1^2+2 \left( \dfrac{ v- v_1}{2}\right)^2$$
$$v^2= v_1^2+ \dfrac{ 2}{4} \left( v^2-2vv_1+ v_1^2 \right) $$
$$v^2= v_1^2+ 0.5 v^2- vv_1+ 0.5 v_1^2 $$
$$0= 1.5v_1^2- 0.5 v^2- vv_1 $$
Plugging from (1),
$$ 1.5v_1^2- 0.5 (7.67)^2- 7.67v_1 =0 $$
$$ 1.5v_1^2- 7.67v_1- 0.5 (7.67)^2 =0 $$
So, the rebounds velocity is $v_1=-2.56\;\rm m/s$ or $v=7.67\;\rm m/s$. The first root is the right speed since it has a negative sign that indicates a negative direction as expected.
The kinetic energy of this package will be completely converted to gravitational potential energy.
Thus,
$$\frac{1}{2}\color{red}{\bf\not} mv_1^2=\color{red}{\bf\not} mgy_f$$
$$\frac{1}{2} v_1^2= gy_f$$
$$\frac{1}{2g} v_1^2= y_f$$
$$ y_f=\dfrac{v_1^2 }{2g}=\dfrac{(-2.56)^2 }{2\times 9.8} =\color{red}{\bf 0.33}\;\rm m$$
