Let $n$ be a fixed natural number.
[b]a)[/b] Find all solutions to the following equation :
\[ \sum_{k=1}^n [\frac x{2^k}]=x-1  \]
[b]b)[/b] Find the number of solutions to the following equation ($m$ is a fixed natural) :
\[ \sum_{k=1}^n [\frac x{2^k}]=x-m  \]
### Part (a)
We need to find all solutions to the equation:
\[ \sum_{k=1}^n \left\lfloor \frac{x}{2^k} \right\rfloor = x - 1 \]

1. **Using the known fact**:
   \[
   \sum_{j \ge 1} \left\lfloor \frac{x}{2^j} \right\rfloor = x - \text{sum of digits of } x \text{ when written in base } 2
   \]
   This implies:
   \[
   \sum_{k=1}^n \left\lfloor \frac{x}{2^k} \right\rfloor = x - \text{sum of the first } n \text{ digits of } x \text{ in base } 2
   \]

2. **Given equation**:
   \[
   x - \text{sum of the first } n \text{ digits of } x \text{ in base } 2 = x - 1
   \]
   This simplifies to:
   \[
   \text{sum of the first } n \text{ digits of } x \text{ in base } 2 = 1
   \]

3. **Conclusion**:
   The sum of the first \( n \) digits of \( x \) in base 2 being 1 means that \( x \) must be a power of 2, specifically \( 2^0, 2^1, \ldots, 2^n \).

Thus, the solutions to part (a) are:
\[
\boxed{2^0, 2^1, \ldots, 2^n}
\]

### Part (b)
We need to find the number of solutions to the equation:
\[ \sum_{k=1}^n \left\lfloor \frac{x}{2^k} \right\rfloor = x - m \]

1. **Express \( x \) in binary form**:
   \[
   x = \sum_{i=0}^{n-1} 2^i a_i + 2^n l
   \]
   where \( a_i \in \{0, 1\} \) and \( l \in \mathbb{Z}_{\ge 0} \).

2. **Using the known fact**:
   \[
   \sum_{k=1}^n \left\lfloor \frac{y}{2^k} \right\rfloor = y - \sum_{i=0}^{n-1} a_i
   \]
   where \( y = \sum_{i=0}^{n-1} 2^i a_i \).

3. **Compute the sum for \( z \)**:
   \[
   \sum_{k=1}^n \left\lfloor \frac{z}{2^k} \right\rfloor = z \left(1 - \frac{1}{2^n}\right) = z - l
   \]
   where \( z = 2^n l \).

4. **Combine the results**:
   \[
   \sum_{k=1}^n \left\lfloor \frac{x}{2^k} \right\rfloor = \left(y - \sum_{i=0}^{n-1} a_i\right) + (z - l) = x - \left(l + \sum_{i=0}^{n-1} a_i\right)
   \]

5. **Given equation**:
   \[
   x - \left(l + \sum_{i=0}^{n-1} a_i\right) = x - m
   \]
   This simplifies to:
   \[
   l + \sum_{i=0}^{n-1} a_i = m
   \]

6. **Count the solutions**:
   For a fixed \( l \) and \( m \), the number of solutions is:
   \[
   \binom{n}{m-l}
   \]
   Summing over all possible values of \( l \):
   \[
   \sum_{l \ge 0} \binom{n}{m-l}
   \]

7. **Final result**:
   The total number of solutions is:
   \[
   \sum_{i=0}^m \binom{n}{i}
   \]

Thus, the number of solutions to part (b) is:
\[
\boxed{\sum_{i=0}^m \binom{n}{i}}
\]