For any positive integer $ k$, let $ f_k$ be the number of elements in the set $ \{ k \plus{} 1, k \plus{} 2, \ldots, 2k\}$ whose base 2 representation contains exactly three 1s. 

(a) Prove that for any positive integer $ m$, there exists at least one positive integer $ k$ such that $ f(k) \equal{} m$.

(b) Determine all positive integers $ m$ for which there exists [i]exactly one[/i] $ k$ with $ f(k) \equal{} m$.
### Part (a)

1. **Define the function \( f_k \)**:
   - \( f_k \) is the number of elements in the set \( \{ k + 1, k + 2, \ldots, 2k \} \) whose base 2 representation contains exactly three 1s.

2. **Understand the range of \( f_k \)**:
   - As \( k \) increases, the set \( \{ k + 1, k + 2, \ldots, 2k \} \) also increases in size.
   - The function \( f_k \) counts the number of elements in this set with exactly three 1s in their binary representation.

3. **Show that \( f_k \) is non-decreasing**:
   - For any \( k \), the set \( \{ k + 1, k + 2, \ldots, 2k \} \) contains \( k \) elements.
   - When \( k \) increases to \( k+1 \), the new set \( \{ k + 2, k + 3, \ldots, 2k + 2 \} \) contains \( k+1 \) elements.
   - The number of elements with exactly three 1s in their binary representation can only increase or stay the same, hence \( f_k \) is non-decreasing.

4. **Prove that \( f_k \) takes all positive integer values**:
   - Since \( f_k \) is non-decreasing and increases by at most 1 at each step, starting from \( f(4) = 1 \), it will eventually take every positive integer value.
   - Therefore, for any positive integer \( m \), there exists a positive integer \( k \) such that \( f(k) = m \).

\(\blacksquare\)

### Part (b)

1. **Condition for exactly one \( k \) with \( f(k) = m \)**:
   - We need to find \( k \) such that \( f(k) = m \) and there is no other \( k' \neq k \) with \( f(k') = m \).

2. **Analyze the binary representation**:
   - For \( f(k) = m \) to be unique, both \( 2k-1 \) and \( 2k+1 \) must have exactly three 1s in their binary representation.

3. **Formulate the condition**:
   - If \( k = 2 + 2^a \) for \( a \geq 2 \), then \( 2k-1 \) and \( 2k+1 \) have exactly three 1s in their binary representation.
   - This is because \( 2k-1 = 4 + 2^{a+1} - 1 \) and \( 2k+1 = 4 + 2^{a+1} + 1 \) both have three 1s.

4. **Determine the value of \( m \)**:
   - The set \( \{ k+1, \ldots, 2k \} \) becomes \( \{ 1 + 2 + 2^a, \ldots, 4 + 2^{a+1} \} \).
   - The number of elements with exactly three 1s in their binary representation is \( \frac{a(a-1)}{2} \).
   - Therefore, \( m = f(2^a) + 1 = \frac{a(a-1)}{2} + 1 \).

5. **Conclusion**:
   - The positive integers \( m \) for which there exists exactly one \( k \) with \( f(k) = m \) are \( m = \frac{a(a-1)}{2} + 1 \) for \( a \geq 2 \).

\(\blacksquare\)

The final answer is \( \boxed{ m = \frac{a(a-1)}{2} + 1 } \) for \( a \geq 2 \).