The problem consists of two parts that require finding a recurrence relation and an analytical solution, followed by proving the correctness via mathematical induction. Let's analyze and solve them one by one.

Part (a):

Problem: $n$ distinct points are marked on a straight line. Into how many parts is the line divided?

1. Recurrence Relation: Each new point added on the line divides it into one additional segment. So, if we denote by $P(n)$ the number of parts into which the line is divided by $n$ points, we can form the recurrence relation: $P(n) = P(n-1) + 1$ where $P(0) = 1$ (since no points leave the line undivided).

2. Analytical Solution: From the recurrence relation, we can see that the number of parts is simply: $P(n) = n + 1$

3. Proof by Induction: Base case: For $n = 0$, the line is divided into 1 part, so $P(0) = 1$, which is correct.

   Inductive step: Assume that for $n = k$, the number of parts is $P(k) = k + 1$. Now, for $n = k + 1$, adding an additional point divides the line into one more part, so $P(k+1) = P(k) + 1 = (k + 1) + 1 = k + 2$, which confirms the inductive hypothesis.

   Hence, by induction, the solution is correct: $P(n) = n + 1$.

Part (b):

Problem: $n$ lines are drawn on a plane such that no two lines are parallel, and no three lines intersect at a single point. Into how many regions is the plane divided?

1. Recurrence Relation: Each new line divides all previously created regions it intersects. Denote by $R(n)$ the number of regions formed by $n$ lines. We can observe that: $R(n) = R(n-1) + n$ where $R(0) = 1$ (since without any lines, the plane is undivided).

2. Analytical Solution: We can solve the recurrence relation by expanding it: $R(n) = R(0) + 1 + 2 + 3 + \dots + n = 1 + \frac{n(n+1)}{2}$ Hence, the analytical formula is: $R(n) = 1 + \frac{n(n+1)}{2}$

3. Proof by Induction: Base case: For $n = 0$, there is 1 region, which matches the formula $R(0) = 1 + \frac{0(0+1)}{2} = 1$.

   Inductive step: Assume that for $n = k$, the number of regions is $R(k) = 1 + \frac{k(k+1)}{2}$. Now, for $n = k + 1$, the new line intersects $k + 1$ existing regions, so the new number of regions is: $R(k+1) = R(k) + (k+1) = \left( 1 + \frac{k(k+1)}{2} \right) + (k+1) = 1 + \frac{k(k+1)}{2} + \frac{2(k+1)}{2} = 1 + \frac{(k+1)(k+2)}{2}$ Thus, the inductive hypothesis holds, and the formula is correct: $R(n) = 1 + \frac{n(n+1)}{2}$.

Would you like more details on any part or a further breakdown of the concepts? Here are five related questions:

1. How would the recurrence relation change if some of the lines in part (b) were parallel?
2. Can we generalize the solution to part (b) if the number of intersections at a point is greater than 3?
3. What is the significance of the induction step in ensuring the validity of a recurrence relation?
4. How would the analytical solution to part (b) change if the lines formed polygons instead of dividing the plane into simple regions?
5. Could this problem be extended to three-dimensional space? How would that change the recurrence relations?

Tip: When solving problems involving recurrence relations, always look for the base case and ensure the relationship between successive terms is clearly understood before attempting an analytical solution.