For question 1, Alice computes $k_0 = B^a, k_1 = (B/A)^a$, and Bob computes a key $k = A^b$. If $c=0$, then $B = g^b$, hence $k_0 = B^a = g^{ab} = A^b = k$; else, if $c=1$, then $B = Ag^b$, hence $k_1 = (B/A)^a = g^{ab} = A^b = k$. In any case, both players have a shared key (either $(k_0,k)$ or $(k_1,k)$).
For question 2, let me give you some hints. Bob generates $B$ according to one of two distributions: (I write $b \gets_{\mathsf{R}} \mathbb{Z}_p$ to indicate that $b$ is taken uniformly at random over $\mathbb{Z}_p$), $D_0 = \{b \gets_{\mathsf{R}}\mathbb{Z}_p: B \gets g^b\}$ and $D_1 = \{b \gets_{\mathsf{R}}\mathbb{Z}_p: B \gets Ag^b\}$. What can you tell about those two distributions? Why cannot one distinguish them, even with unlimited computational power? Additional hint: $A = g^a$; what can be said about the distribution of exponents $a+b \bmod p$ when $b$ is taken at random from $\mathbb{Z}_p$, but $a$ is fixed?
For question 3, let us suppose that there is a player, Bob, that is able to compute both $k_0$ and $k_1$ with some fixed probability $\epsilon$ after performing a one-out-of-two key exchange with Alice. Suppose now that you are given a challenge for the computational Diffie-Hellman problem: you receive the description of a group $\mathbb{G}$, as well as a tuple $(g, X= g^x, Y = g^y)$. You break the CDH assumption if you manage to return $Z = g^{xy}$ with some non-negligible probability. Therefore, your goal is now to construct a $Z$ which will have a good probability (related to $\epsilon$) to be equal to $g^{xy}$, by interacting with Bob: you'll play a one-of-of-two key exchange with him, using carefully chosen elements from the CDH challenge instead of picking group elements according to the specification of the key exchange. Then, you'll reconstruct $g^{xy}$ with good probability using the keys $k_0,k_1$ that Bob returns (recall that they have a probability $\epsilon$ to be the correct keys).
I've given many details, so you should be able to figure out how to properly answer the question by now. If you are still blocked, if there is something you do not understand, please do not hesitate to mention it and I'll provide further insights.
EDIT: further insights for question 3.
Suppose now you received $(g, X= g^x, Y = g^y)$. We will play Alice's role in a one-out-of-two key exchange. As you remarked in your comment, we cannot gain anything from the values $B=g^b$ chosen by Bob - as it is chosen by the adversary, he can always compute it. However, we can gain something from the fact that Bob will compute both $k_0 = B^a$ and $k_1 = (B/A)^a$: we can deduce $A^a$ from that by computing $k_0/k_1$. Note that $A^a = g^{a^2}$.
Let me sum up: we have an adversary Bob, to which we can send some $A = g^a$ and who eventually returns two keys. At the end of the interaction with Bob, by computing $k_0/k_1$, we recover $g^{a^2}$ (with probability at least $\epsilon$, as this is the probability that both keys are correct).
So, we can now see Bob as an oracle which returns $g^{a^2}$ when we send it $g^{a}$. Given such an oracle, solving the CDH problem is a standard exercise. Can you see how it is done? The trick is simply: use the identity $ab = 1/2((a+b)^2 - a^2 - b^2)$. Now, from the "exponent squaring oracle" and $g^x,g^y$, you should be able to find out $g^{xy}$.
If you still do not see how to solve the exercise, I'll give you the entire solution, but it's way more interesting and useful for you if you figure it out for yourself.
