To gain an appreciation for how easily we can be led astray through flawed reasoning, consider the following "intuitive" approach:
Incorrect: Among all boxes, there are $3$ gold coins and $3$ silver coins. Given that a gold coin was obtained in the first draw, this means that either the first or last box was chosen. In only one of these two cases is the other coin gold, therefore the answer is $1/2$.
Now, before we proceed with the correct line of reasoning, ask yourself why the above reasoning is flawed.
Correct: Label the coins in each box as follows: $$\begin{array}{ccc}
\text{Box 1}& \text{Box 2} & \text{Box 3} \\
\hline \{g_1, g_2\} & \{s_1, s_2\} & \{g_3, s_3\} \end{array}$$ Then the act of selecting a box at random and drawing one of the two coins in that box, followed by drawing the second coin in that same box, yields the possible outcomes $$(g_1, g_2), (g_2, g_1), \\
(s_1, s_2), (s_2, s_1), \\
(g_3, s_3), (s_3, g_3).$$
Unlike in our previous use of notation, notice here that the order matters. Each of the six outcomes is equally likely, and for instance, $(g_1, g_2)$ is distinct from $(g_2, g_1)$ even though both coins are gold, because you have labeled the coins.
Then we can immediately see that $3$ of the outcomes have $g$ in the first position: $$(g_1, g_2), (g_2, g_1), (g_3, s_3).$$ And we can see that in exactly two of the three outcomes, $g$ is also in the second position. Therefore, the probability is $2/3$.
You have made a calculation error. We must have $$\Pr[A \mid B] + \Pr[A^c \mid B] = 1.$$ The second probability should be $\Pr[A^c \mid B] = 8/11$.
The rest of the question is straightforward: the probability that the second coin drawn from the same box is also silver is equal to $$0 \Pr[A \mid B] + (1/2) \Pr[A^c \mid B] = 4/11.$$ This is because if the box that was chosen was the first one, there are no more silver coins to be drawn. If the box was the second one, then of the remaining two coins in the box, one is silver and will be drawn with probability $1/2.$
Best Answer
Here's a more intuitive answer.
You have two random things going on: your choice of coin is random, and the side facing up is random. So in fact you're picking one of 8 coin-sides out of the bag, and each one occurs with equal weighting.
So let's label the sides: G1 and G2, S1 and S2, G3 and S3, and G4 and S4 (where G = gold, S = silver, and they happen to be fused together as I have written them). You choose a random coin-side. It's gold, so, it must be G1, G2, G3 or G4 - all equally probable. The reverses of those coins are, in order, G2, G1, S3 or S4 - all still equally probable, of course. So the probability is 1/2.
This simply corresponds to the fact that, if you pick a random coin-side out of the bag and it's gold, it's twice as likely to be the G1-G2 coin than the G3-S3 coin (because that coin has twice as many gold sides).