a.) For all positive integer $k$ find the smallest positive integer $f(k)$ such that $5$ sets $s_1,s_2, \ldots , s_5$ exist satisfying:
 
[b]i.[/b] each has $k$ elements;
[b]ii.[/b] $s_i$ and $s_{i+1}$ are disjoint for $i=1,2,...,5$ ($s_6=s_1$)
[b]iii.[/b] the union of the $5$ sets has exactly $f(k)$ elements.

b.) Generalisation: Consider $n \geq 3$ sets instead of $5$.
### Part (a)

1. **Understanding the Problem:**
   We need to find the smallest positive integer \( f(k) \) such that there exist 5 sets \( s_1, s_2, \ldots, s_5 \) each with \( k \) elements, where \( s_i \) and \( s_{i+1} \) are disjoint for \( i = 1, 2, \ldots, 5 \) (with \( s_6 = s_1 \)), and the union of these 5 sets has exactly \( f(k) \) elements.

2. **Initial Observation:**
   If \( n \) is even, then \( f_n(k) = 2k \) for all \( k \). This is because we can pair the sets such that each pair is disjoint and covers all elements.

3. **Claim for Odd \( n \):**
   We claim that for odd \( n \), specifically \( n = 5 \), \( f_5(k) = \left\lceil \frac{5k}{2} \right\rceil \).

4. **Proof:**
   - Consider the indices \( i \) in \( s_i \) modulo 5, and the elements as \( 1, 2, \ldots, k \) modulo \( k \).
   - Suppose we have fewer than \( \frac{5k}{2} \) elements. For any \( i \), \( s_{i-1} \) and \( s_{i+1} \) must both be disjoint from \( s_i \), so they must share more than \( \frac{3k}{2} \) elements.
   - Applying this to \( s_1, s_3, \ldots, s_5 \), we have that \( s_1 \) and \( s_5 \) must share more than \( \frac{k}{2} \) elements.
   - Since \( s_0 \) and \( s_1 \) are disjoint, \( s_0 \) and \( s_{2i-1} \) share fewer than \( \frac{3k}{2} \) elements. This leaves fewer than \( k \) elements for \( s_5 \), which must be disjoint from both.

5. **Constructing the Sets:**
   - Let there be \( m = \left\lceil \frac{5k}{2} \right\rceil \) elements.
   - We can satisfy the conditions by letting \( s_{2i} = \left\{ \left\lfloor \frac{im}{5} \right\rfloor + j \mid j = 1, 2, \ldots, k \right\} \).
   - For any \( i \), we have:
     \[
     s_{2i} = \left\{ \left\lfloor \frac{im}{5} \right\rfloor + j \mid j = 1, 2, \ldots, k \right\}
     \]
     \[
     s_{2i+1} = s_{2i+2n+2} = \left\{ \left\lfloor \frac{im + nm + m}{5} \right\rfloor + j \mid j = 1, 2, \ldots, k \right\}
     \]

6. **Verification:**
   - Since:
     \[
     \left\lfloor \frac{im + nm + m}{5} \right\rfloor - \left\lfloor \frac{im}{5} \right\rfloor \geq \left\lfloor \frac{(n+1)m}{5} \right\rfloor \geq k
     \]
     and
     \[
     \left\lfloor \frac{im + nm + m}{5} \right\rfloor - \left\lfloor \frac{im}{5} \right\rfloor \leq \left\lceil \frac{(n+1)k}{n} \right\rceil = m - k
     \]
   - We see that \( s_{2i} \) and \( s_{2i+1} \) are disjoint. Since \( s_{2i} \) covers all \( s_j \) for \( i = 1, 2, \ldots, 5 \), we are done.

### Part (b)

1. **Generalization:**
   Consider \( n \geq 3 \) sets instead of 5.

2. **Even \( n \):**
   If \( n \) is even, \( f_n(k) = 2k \) for all \( k \).

3. **Odd \( n \):**
   We claim that \( f_{2n+1}(k) = \left\lceil \frac{(2n+1)k}{n} \right\rceil \).

4. **Proof:**
   - Similar to the proof for \( n = 5 \), we can generalize the construction and verification steps for any odd \( n \).

The final answer is \( \boxed{ f_{2n+1}(k) = \left\lceil \frac{(2n+1)k}{n} \right\rceil } \).