The answer to the first question is correct.
What is the initial probability p(nS) that the bag contains nS silver coins?
$$P(nS) = 1/11 $$
Now, the second question
You pull out one coin: it is silver. What is now the probability that the bag contains 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 silver coins?
The analysis in the question is wrong because the question is about conditional probability. We want to find out the probability of the event that the bag contains n silver coins($0 \le n \le 9$), given that the first coin pulled out was a silver coin. We will use Bayes' theorem to solve this.
See this and this for Bayes theorem and this for an illustrative example. The idea of drawing all the probabilities as a tree demonstrated in this idea is quite helpful in understanding the problem. Now, solution to the problem.
Lets us call the event we pulled out one silver coin as $O$.
Let us call the event there are n silver coins remaining in the bag as $nS$, where $0 \le n \le 9$
We want to find out $P(nS | O)$
Using Bayes formula
$$P(nS|O) = \frac{P(O|nS). P(nS)}{P(O)}$$
From the above question, we know that
$$P(nS) = 1/11 $$
$P(O|nS)$ is defined as the probability where a silver coin is picked given there are n silver coins remaining in the bag after the ball is picked. Since there are n silver coins remaining after the first silver coin was picked, the probability $P(O|nS)$ can be defined as $\frac{n+1}{10}$
$$P(O|nS) = \frac{n+1}{10}$$
$P(O)$ is the probability that a silver coin is picked from the bag on the first turn. Since there are either silver or gold coins in the bag, there is equal probability for both. Thus,
$$P(O) = 1/2$$
This can also be derived by drawing the tree as illustrated in the video
$$P(O) = (\frac{1}{11}*\frac{0}{10}) + (\frac{1}{11}*\frac{1}{10}) + .... + (\frac{1}{11}*\frac{9}{10}) + (\frac{1}{11}*\frac{10}{10})$$
$$P(O) = \frac{1}{11}(\frac{0}{10} + \frac{1}{10} + .... + \frac{9}{10} + \frac{10}{10})$$
$$P(O) = \frac{1}{11}(\frac{0+1+2+ ... + 9 + 10}{10}) = \frac{55}{11*10} = \frac{1}{2}$$
Substituting these values in
$$P(nS|O) = \frac{P(O|nS). P(nS)}{P(O)}$$
$$P(nS|O) = \frac{n+1}{55}$$
You now pull out another coin: it is gold. What is now the probability that the bag contains 0, 1, 2, 3, 4, 5, 6, 7, 8 silver coins?
Let us call the event where two coins were pulled, and the first one is silver and other is gold as $T$.
Let us call the event there are n silver coins remaining in the bag as $nS$, where $0 \le n \le 8$
We want to find out $P(nS | T)$
Using Bayes formula
$$P(nS|T) = \frac{P(T|nS). P(nS)}{P(T)}$$
We know that
$$P(nS) = 1/11 $$
$P(T|nS)$ is defined as the probability where a silver coin is drawn first and gold coin is drawn next given there are n silver coins remaining in the bag after the two coins are picked. On the first turn, the probability for a silver coin to be picked is $\frac{n+1}{10}$. In the second turn, 9 coins would be left in the bag and n of them are silver coins, so, there would be $9-n$ gold coins. The probability of picking a gold coin in the second turn would be $\frac{9-n}{9}$. Combined, the probability would be
$$P(T|nS) = \frac{n+1}{10}*\frac{9-n}{9}$$
$P(T)$ is the probability that a black ball is picked from the bag first followed by a white ball. This can be derived from the probability tree
$$P(T) = (\frac{1}{11}*\frac{0}{10}) + (\frac{1}{11}*\frac{1}{10}*\frac{9}{9}) + (\frac{1}{11}*\frac{2}{10}*\frac{8}{9}) .... + (\frac{1}{11}*\frac{9}{10}*\frac{1}{9}) + (\frac{1}{11}*\frac{10}{10}*\frac{0}{9})$$
$$P(T) = \frac{1}{11}(\sum_{n=0}^{10} \frac{n(10-n)}{10*9})$$
$$P(T) = \frac{1}{11}(\frac{165}{90}) = \frac{1}{6}$$
Substituting these values in
$$P(nS|T) = \frac{P(T|nS). P(nS)}{P(T)}$$
$$P(nS|T) = \frac{8n+9-n^2}{165}$$
Best Answer
Use a combinatorial approach and assume that order doesn't matter. Then for the first question, we obtain: $$ \frac{\binom{6}{1}\binom{4}{1}\binom{3}{1}}{\binom{13}{3}} = \frac{6 \cdot 4 \cdot 3}{\frac{13 \cdot 12 \cdot 11}{3 \cdot 2}} = \frac{6 \cdot 4 \cdot 3}{13 \cdot 2 \cdot 11} = \frac{36}{143} $$ As bof pointed out in the comments, your answer was off by a factor of $3! = 6$, since you implicitly assumed a particular order of the $3$ types of coins.
For the second question, we consider all $3$ of each type of coin as separate cases (which was exactly what you were thinking of doing). This yields: $$ \frac{\binom{6}{3} + \binom{4}{3} + \binom{3}{3}}{\binom{13}{3}} = \frac{\frac{6 \cdot 5 \cdot 4}{3 \cdot 2} + 4 + 1}{\frac{13 \cdot 12 \cdot 11}{3 \cdot 2}} = \frac{5 \cdot 4 + 5}{13 \cdot 2 \cdot 11} = \frac{25}{286} $$