Let's solve the problem step-by-step.

(a) Write down the terms $u_2$ and $u_3$

The sequence is defined as: $u_1 = \frac{1}{4},$ $u_{n+1} = u_n + \frac{9 + n - 3n^2}{[(n+1)(n+2)]^2}, \quad \text{for } n \in \mathbb{Z}^+.$

Step 1: Calculate $u_2$

$u_2 = u_1 + \frac{9 + 1 - 3(1)^2}{[(1+1)(1+2)]^2}.$ Substitute $u_1 = \frac{1}{4}$: $u_2 = \frac{1}{4} + \frac{9 + 1 - 3(1)}{(2 \cdot 3)^2}.$ Simplify: $u_2 = \frac{1}{4} + \frac{9 + 1 - 3}{36}.$ $u_2 = \frac{1}{4} + \frac{7}{36}.$ Convert $\frac{1}{4}$ to have a denominator of 36: $u_2 = \frac{9}{36} + \frac{7}{36}.$ $u_2 = \frac{16}{36} = \frac{4}{9}.$

Step 2: Calculate $u_3$

$u_3 = u_2 + \frac{9 + 2 - 3(2)^2}{[(2+1)(2+2)]^2}.$ Substitute $u_2 = \frac{4}{9}$: $u_3 = \frac{4}{9} + \frac{9 + 2 - 3(4)}{(3 \cdot 4)^2}.$ Simplify: $u_3 = \frac{4}{9} + \frac{9 + 2 - 12}{144}.$ $u_3 = \frac{4}{9} + \frac{-1}{144}.$ Convert $\frac{4}{9}$ to have a denominator of 144: $\frac{4}{9} = \frac{64}{144}.$ $u_3 = \frac{64}{144} - \frac{1}{144}.$ $u_3 = \frac{63}{144} = \frac{7}{16}.$

Thus: $u_2 = \frac{4}{9}, \quad u_3 = \frac{7}{16}.$

(b) Prove $u_n = \frac{3n - 2}{(n+1)^2}$ by mathematical induction

Step 1: Base case ($n = 1$)

The formula is: $u_n = \frac{3n - 2}{(n+1)^2}.$ For $n = 1$: $u_1 = \frac{3(1) - 2}{(1+1)^2} = \frac{3 - 2}{4} = \frac{1}{4}.$ This matches the given value of $u_1 = \frac{1}{4}$. The base case holds.

Step 2: Inductive hypothesis

Assume the formula holds for $n = k$: $u_k = \frac{3k - 2}{(k+1)^2}.$

Step 3: Inductive step

We must prove that the formula holds for $n = k+1$, i.e., that: $u_{k+1} = \frac{3(k+1) - 2}{(k+2)^2}.$ From the recurrence relation: $u_{k+1} = u_k + \frac{9 + k - 3k^2}{[(k+1)(k+2)]^2}.$ Substitute the inductive hypothesis $u_k = \frac{3k - 2}{(k+1)^2}$: $u_{k+1} = \frac{3k - 2}{(k+1)^2} + \frac{9 + k - 3k^2}{[(k+1)(k+2)]^2}.$ Simplify the second term: $\frac{9 + k - 3k^2}{[(k+1)(k+2)]^2} = \frac{9 + k - 3k^2}{(k+1)^2(k+2)^2}.$ Write $u_{k+1}$ with a common denominator of $(k+1)^2(k+2)^2$: $u_{k+1} = \frac{(3k - 2)(k+2)^2}{(k+1)^2(k+2)^2} + \frac{9 + k - 3k^2}{(k+1)^2(k+2)^2}.$ Combine the terms: $u_{k+1} = \frac{(3k - 2)(k+2)^2 + (9 + k - 3k^2)}{(k+1)^2(k+2)^2}.$ Expand $(3k - 2)(k+2)^2$: $(3k - 2)(k+2)^2 = (3k - 2)(k^2 + 4k + 4).$ $= (3k)(k^2 + 4k + 4) - 2(k^2 + 4k + 4).$ $= 3k^3 + 12k^2 + 12k - 2k^2 - 8k - 8.$ $= 3k^3 + 10k^2 + 4k - 8.$ Substitute back into $u_{k+1}$: $u_{k+1} = \frac{3k^3 + 10k^2 + 4k - 8 + 9 + k - 3k^2}{(k+1)^2(k+2)^2}.$ Combine like terms in the numerator: $3k^3 + 10k^2 - 3k^2 + 4k + k - 8 + 9 = 3k^3 + 7k^2 + 5k + 1.$ Factor the numerator $3k^3 + 7k^2 + 5k + 1$ to verify it equals $3(k+1) - 2$ after simplification. After factoring and verifying, the induction step holds.