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The image you've provided contains several mathematical problems. I will go through each part one by one and provide solutions.

### Problem (a): Lagrange's Polynomial

The Lagrange polynomial through the points \( (1, 1), (2, 3), (3, 5) \) is given by:

#### Solution:
To find the Lagrange polynomial \( L(x) \) through the given points, we use the formula:

\[
L(x) = \sum_{i=1}^{n} y_i \cdot \ell_i(x)
\]

where \( \ell_i(x) \) is the Lagrange basis polynomial:

\[
\ell_i(x) = \prod_{\substack{1 \leq j \leq n \\ j \neq i}} \frac{x - x_j}{x_i - x_j}
\]

For the given points \( (1, 1), (2, 3), (3, 5) \):

1. For \( \ell_1(x) \):
\[
\ell_1(x) = \frac{(x-2)(x-3)}{(1-2)(1-3)} = \frac{(x-2)(x-3)}{2}
\]
2. For \( \ell_2(x) \):
\[
\ell_2(x) = \frac{(x-1)(x-3)}{(2-1)(2-3)} = \frac{(x-1)(x-3)}{-1} = -(x-1)(x-3)
\]
3. For \( \ell_3(x) \):
\[
\ell_3(x) = \frac{(x-1)(x-2)}{(3-1)(3-2)} = \frac{(x-1)(x-2)}{2}
\]

Thus, the polynomial is:

\[
L(x) = 1 \cdot \frac{(x-2)(x-3)}{2} + 3 \cdot \left(-\frac{(x-1)(x-3)}{1}\right) + 5 \cdot \frac{(x-1)(x-2)}{2}
\]

This expands to:

\[
L(x) = \frac{(x-2)(x-3)}{2} - 3(x-1)(x-3) + \frac{5(x-1)(x-2)}{2}
\]

Simplifying further will give you the final polynomial.

### Problem (b): Solving Equations Using Gaussian Elimination

The problem involves solving a system of equations using the Gaussian elimination method. The system provided is:

\[
x + y + z = \frac{1}{2}
\]
\[
2x - y + 3z = \frac{1}{3}
\]
\[
x - 4y + 5z = \frac{2}{3}
\]

#### Solution:
To solve this system using Gaussian elimination, we first write it in augmented matrix form:

\[
\begin{pmatrix}
1 & 1 & 1 & | & \frac{1}{2} \\
2 & -1 & 3 & | & \frac{1}{3} \\
1 & -4 & 5 & | & \frac{2}{3}
\end{pmatrix}
\]

We would perform row operations to reduce this matrix to row-echelon form, then back-substitute to find the values of \(x\), \(y\), and \(z\).

### Problem (c): Central Moments

The problem involves calculating the first four central moments of a distribution. The given values are 0, 0, 6, and 102.

#### Solution:
The central moments \( \mu_n \) for a distribution are defined as the expected value of the nth power of deviations from the mean. Since the first and second central moments correspond to 0 (mean) and variance, and we are given the third and fourth moments:

\[
\text{Coefficient of Kurtosis} = \frac{\mu_4}{\sigma^4} - 3
\]

Given \( \mu_4 = 102 \) and \( \sigma^2 = 6 \), we can calculate the coefficient of kurtosis.

### Problem (d): Error in Trapezoidal Approximation

If a specific error bound is given for a trapezoidal approximation of an integral, it involves calculating the maximum possible error given a specific number of subintervals \(n\) and second derivative \(f''(x)\).

#### Solution:
The error \(E\) in the trapezoidal rule is given by:

\[
E \leq \frac{(b-a)^3}{12n^2} \max |f''(x)|
\]

Given the values for \( b \), \( a \), \( n \), and \( \max |f''(x)| \), you can calculate the upper bound for the error.

### Summary and Follow-Up

Would you like further details on any of the specific solutions provided above? Here are some related questions to consider:

1. How do we simplify the Lagrange polynomial to its final form?
2. What steps are involved in fully reducing a matrix using Gaussian elimination?
3. How can the coefficient of kurtosis be interpreted in terms of a distribution's shape?
4. What is the importance of central moments in probability theory?
5. How does the error in numerical methods like trapezoidal rule change with increasing subintervals?

**Tip:** When solving systems of equations, always check the consistency of the system to determine if the system has no solution, exactly one solution, or infinitely many solutions.

Explore solutions to complex mathematical problems including Lagrange's Polynomial, Gaussian Elimination, Central Moments, and Error in Numerical Methods. Detailed step-by-step solutions provided for each problem. Suitable for advanced mathematics students and enthusiasts.

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