Answer
See the detailed answer.
Work Step by Step
a) Since the surface is frictionless, we can consider the sled and the Earth an isolated system.
This means that the energy is conserved.
$$E_i=E_f$$
where $E_i$ is when the sled is at the highest point (at which $\phi=0$), and $E_f$ is when the angle is $\phi$, as we see in the figures below.
$$K_i+U_{gi}=K_f+U_{gf}$$
The sled starts from rest, so its initial kinetic energy is zero.
$$0+U_{gi}=K_f+U_{gf}$$
$$0+\color{red}{\bf\not}mgy_i=\frac{1}{2}\color{red}{\bf\not}mv^2+\color{red}{\bf\not}mgy_f$$
where $y_i=R$ and $y_f=h$
Thus,
$$\frac{1}{2}v^2=gR-gh=g(R-h)$$
where $h$ is given by $\cos\phi =\dfrac{h}{R}$, so $h=R\cos\phi $
$$v^2 =2g(R-R\cos\phi )$$
$$\boxed{v =\sqrt{2gR(1- \cos\phi )}}$$
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b)
As we see in the second figure below, we chose the tangential direction to be our positive $x$-direction (which changes with time), and the perpendicular to it is the positive $r$-direction.
We need to analyze the forces exerted on the $x$-direction and apply Newton's second law.
$$\sum F_r= F_n-mg\cos\phi= ma_r$$
Hence,
$$ F_n-mg\cos\phi=a_r=\dfrac{v^2}{R}$$
Thus,
$$v^2=\dfrac{R(F_n-mg\cos\phi)}{m}$$
The maximum speed without leaving the surface at $\phi$ occurs when the sled is barely touching the surface.
This means that the normal force is closer to zero, $F_n\rightarrow0$.
So,
$$v_{max}^2=\dfrac{R(0-mg\cos\phi)}{m}$$
Therefore,
$$v_{max} =\sqrt{\dfrac{ \color{red}{\bf\not}mgR\cos\phi }{\color{red}{\bf\not}m}}$$
$$\boxed{v_{max} =\sqrt{ gR\cos\phi }}$$
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c) When the speed of the sled reaches the maximum speed (which we found above in part b), the sled will fly off at an angle of $\phi_{max}$.
From the two boxed formulas;
$$\sqrt{2gR(1- \cos\phi )}=\sqrt{ gR\cos\phi }$$
Squaring both sides;
$$2\color{red}{\bf\not}g\color{red}{\bf\not}R(1- \cos\phi )= \color{red}{\bf\not}g\color{red}{\bf\not}R\cos\phi $$
Hence,
$$\cos\phi=\dfrac{2}{3}$$
Therefore,
$$\phi_{max}=\color{red}{\bf 48.2^\circ}$$
