Answer
a) $2.25\;\rm m/s$
b) $v_\theta = \sqrt{ 9-1.96(1-\cos\theta) }$
Work Step by Step
a) We can assume that the ice cube is sliding inside the pipe since it is very slippery. We choose the system to be the cube plus the earth. This is an isolated system so the energy is conserved.
Thus,
$$E_i=E_f$$
$$K_{1}+U_{g1}=K_{2}+U_{g2}$$
where $1$ is the initial position of the cube at the highest point in its vertical circular path, and $2$ is the final position of the cube at the bottom.
$$\frac{1}{2}\color{red}{\bf\not} mv_1^2+ \color{red}{\bf\not} mgy_1=\frac{1}{2}\color{red}{\bf\not} mv_2^2+\overbrace{mgy_2}^{=0}$$
As we see in the figure below, $y_1=2R_{pipe}=D_{pipe}$ where $D$ is the diameter of the pipe, and $y_2=0$.
$$\frac{1}{2} v_1^2+ gD_{pipe}=\frac{1}{2} v_2^2 \tag 1$$
Solving for $v_1$; which is the cube's speed at the top
$$v_1^2=2\left[\frac{1}{2} v_2^2-gD_{pipe}\right]$$
$$v_1 = \sqrt{v_2^2-2gD_{pipe}}$$
Plugging the known;
$$v_1 = \sqrt{3^2-(2\times 9.8\times 0.2) }=\color{red}{\bf 2.25}\;\rm m/s$$
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b) From the geometry of the figure below, we can see that the height as a function of $\theta$ is given by $y=R-d$ where $ \cos\theta=\dfrac{d}{R}$, so $d=R\cos\theta$
Hence,
$$y=R-R\cos\theta=R(1-\cos\theta)$$
Plugging that into (1) by replacing $D_{pipe}$ by $y=R(1-\cos\theta)$,
$$\frac{1}{2} v_\theta^2+ gR(1-\cos\theta)=\frac{1}{2} v_2^2 $$
where $v_\theta$ is the speed at angle $\theta$ and $v_2$ the speed at the bottom when $\theta=0$.
Solving for $v_\theta$;
$$ v_\theta^2 = v_2^2-2gR(1-\cos\theta) $$
$$ v_\theta = \sqrt{ v_2^2-2gR(1-\cos\theta) }$$
Plugging the known;
$$ v_\theta = \sqrt{ 3^2-2(9.8)(0.1)(1-\cos\theta) }$$
$$ \boxed{ v_\theta = \sqrt{ 9-1.96(1-\cos\theta) }}$$
At an angle of zero, $v_\theta=3\;\rm m/s$, and at and angle of $180^\circ$, $v_\theta=\sqrt{ 9-1.96(1-(-1)) }=\bf 2.25\;\rm m/s$ which as we found above. So this formula is correct.
