Answer
See the detailed answer below.
Work Step by Step
a) We drew the $U$ versus $r$ graph, as seen below.
For two stars, $r$, which is the distance between their centers, is never de zero since they will collide together in this case and will be treated as one unit. Hence, the hyperbola seen below.
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b) According to the given formula, the zero potential occurs at an infinite distance between our two stars (or two objects).
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$$U=\dfrac{-Gm_1m_2}{r}=\dfrac{-Gm_1m_2}{\infty}=0$$
This makes sense since both stars are not interacting at all when the distance between them is infinite.
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c) At the moment of impact, the separation distance between them will be the sum of their radius.
$$r_f=R_1+R_2\tag1$$
The system here is the two stars only and we assume that the system is isolated and there are no other objects affecting it.
Thus,
$$E_f=E_i$$
$$K_i+U_i=K_f+U_f$$
The two stars are initially at rest, so the initial kinetic energy of the system is zero; $K_i=0$.
$$0+U_i=K_f+U_f$$
$$ \dfrac{-Gm_1m_2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\dfrac{-Gm_1m_2}{r_f}$$
Plugging from (1);
$$ \dfrac{-Gm_1m_2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\dfrac{-Gm_1m_2}{r_f}$$
Thus,
$$ \dfrac{-Gm_1m_2}{r_i}+\dfrac{ Gm_1m_2}{r_f}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$
$$2 Gm_1m_2 \left[ \dfrac{1}{r_f}-\dfrac{ 1}{r_i}\right]= m_1v_1^2+ m_2v_2^2\tag 2$$
We also know that the momentum is conserved since the system is isolated.
Thus,
$$p_i=p_f$$
We know that the two stars were at rest, so the initial linear momentum of the system is zero.
So,
$$0=p_f$$
$$m_1v_1+m_2v_2=0$$
Thus, we can solve for any speed of both of them.
We assume that $v_1$ is toward the right and hence $v_2$ toward the left.
Let's solve for $v_2$;
$$v_2=\dfrac{-m_1v_1}{m_2}\tag 3$$
Plugging $v_2$ from (3) into (2);
$$2 Gm_1m_2 \left[ \dfrac{1}{r_f}-\dfrac{ 1}{r_i}\right]= m_1v_1^2+ m_2 \left(\dfrac{-m_1v_1}{m_2}\right)^2$$
$$2 Gm_1m_2 \left[ \dfrac{1}{r_f}-\dfrac{ 1}{r_i}\right]= m_1v_1^2+ \dfrac{ m_1^2v_1^2}{m_2} $$
$$2 Gm_1m_2 \left[ \dfrac{1}{r_f}-\dfrac{ 1}{r_i}\right]=\left[ m_1+ \dfrac{ m_1^2 }{m_2}\right]v_1^2 $$
Thus,
$$v_1=\sqrt{\dfrac{2 Gm_1m_2 \left[ \dfrac{1}{r_f}-\dfrac{ 1}{r_i}\right]}{m_1+ \dfrac{ m_1^2 }{m_2}}}$$
Plugging from (1);
$$v_1=\sqrt{\dfrac{2 Gm_1m_2 \left[ \dfrac{1}{R_1+R_2}-\dfrac{ 1}{r_i}\right]}{m_1+ \dfrac{ m_1^2 }{m_2}}}$$
Plugging the known;
$$v_1=\sqrt{\dfrac{2 (6.67\times 10^{-11})(2\times 10^{30})(8\times 10^{30})\left[ \dfrac{1}{(7+11)10^8}-\dfrac{ 1}{10^{14}}\right]}{(2\times 10^{30})+ \dfrac{ (2\times 10^{30})^2 }{(8\times 10^{30})}}}$$
$$v_1=\color{red}{\bf 6.89\times 10^5}\;\rm m/s$$
Plugging into (3), and plug the known as well.
$$v_2=\dfrac{- (2\times 10^{30})( 6.89\times 10^5) }{(8\times 10^{30})} $$
$$v_2=\color{red}{\bf -1.72\times 10^5}\;\rm m/s$$
