[b]a)[/b] Among $9$ apparently identical coins, one is false and lighter than the others. How can you discover the fake coin by making $2$ weighing in a two-course balance?
[b]b)[/b] Find the least necessary number of weighing that must be done to cover a false currency between $27$ coins if all the others are true.
### Part (a)
To find the fake coin among 9 coins using only 2 weighings, we can use the following strategy:

1. **First Weighing:**
   - Divide the 9 coins into three groups of 3 coins each: \(A, B, C\).
   - Weigh group \(A\) against group \(B\).

2. **Second Weighing:**
   - If \(A\) and \(B\) are equal, the fake coin is in group \(C\). Weigh two coins from group \(C\) against each other.
     - If they are equal, the remaining coin in group \(C\) is the fake coin.
     - If they are not equal, the lighter one is the fake coin.
   - If \(A\) and \(B\) are not equal, the lighter group contains the fake coin. Weigh two coins from the lighter group against each other.
     - If they are equal, the remaining coin in the lighter group is the fake coin.
     - If they are not equal, the lighter one is the fake coin.

Thus, we can find the fake coin in exactly 2 weighings.

### Part (b)
To find the least number of weighings required to identify a fake coin among 27 coins, we can use the following reasoning:

1. **General Strategy:**
   - Let \(S\) be a set of \(N\) coins. We claim that the smallest number of weighings needed to find the fake coin in \(S\) is \(\lceil \log_3 N \rceil\).
   - Consider the case when \(|S| = 3^n\), where \(n \in \mathbb{N}\). We can split \(S\) into three sets \(S_1, S_2, S_3\), where each has \(3^{n-1}\) coins.
   - Compare the weights of \(S_1\) and \(S_2\):
     - If \(S_1\) and \(S_2\) are equal, then \(S_3\) has the counterfeit coin.
     - Otherwise, one of \(S_1\) or \(S_2\) has the counterfeit coin.
   - This narrows down the search to one set with \(3^{n-1}\) coins. Repeat this procedure until we are left with \(3^0 = 1\) coin, which will be the counterfeit coin.
   - Thus, a total of \(n\) weighings will have been conducted.

2. **Proof of Minimum Weighings:**
   - Suppose we have \(3^{n-1} < |S| < 3^n\). We can split \(S\) into three sets, \(S_1, S_2, S_3\), such that each set has at most \(3^{n-1}\) coins and \(|S_1| = |S_2|\).
   - Compare the weights of \(S_1\) and \(S_2\) to isolate one of the three sets to inspect further as done previously.
   - Repeat this process until we have one coin left, resulting in a total of \(n\) weighings.
   - Therefore, for a set of \(3^{n-1} < |S| \leq 3^n\) coins, we can find the fake coin in \(n\) steps.

Thus, we can find the fake coin in \(\lceil \log_3 |S| \rceil = \lceil \log_3 N \rceil\) weighings.

### Conclusion
For 27 coins, the number of weighings required is:
\[
\lceil \log_3 27 \rceil = \lceil 3 \rceil = 3
\]

The final answer is \( \boxed{ 3 } \) weighings.