The event G can be reinterpreted as Box 3 not chosen
Therefore, $P(GF) =P(G) P(F|G)$.
Here, $P(G) = 2/3$ and $P(F|G) =5/6$.
As you have already calculated the denominator to be $2/3$, therefore final answer is $5/6$
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 $2/3$.
Best Answer
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 $1/2.$