Let's solve the given problem step by step.

1. a) Power Series Expansion of $f(x) = \frac{1 - \cos\frac{x}{2}}{x^2}$

Step 1: Expand $1 - \cos\frac{x}{2}$ into a Taylor series.

Using the Taylor series expansion of $\cos u$ around $u = 0$: $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots$ For $u = \frac{x}{2}$: $\cos\frac{x}{2} = 1 - \frac{\left(\frac{x}{2}\right)^2}{2!} + \frac{\left(\frac{x}{2}\right)^4}{4!} - \dots$ Thus: $1 - \cos\frac{x}{2} = \frac{\left(\frac{x}{2}\right)^2}{2!} - \frac{\left(\frac{x}{2}\right)^4}{4!} + \dots = \frac{x^2}{8} - \frac{x^4}{192} + \dots$

Step 2: Divide by $x^2$.

$f(x) = \frac{1 - \cos\frac{x}{2}}{x^2} = \frac{\frac{x^2}{8} - \frac{x^4}{192} + \dots}{x^2} = \frac{1}{8} - \frac{x^2}{192} + \dots$

The first three terms of the power series expansion are: $f(x) = \frac{1}{8} - \frac{x^2}{192} + \dots$

Step 3: Determine the radius of convergence.

The series converges where the Taylor series for $\cos \frac{x}{2}$ converges, which is for all $x \in \mathbb{R}$, as the series for cosine converges everywhere. Hence, the interval of convergence is $(-\infty, \infty)$.

1. b) Approximate $\int_0^1 g(x) \, dx$, where:

$g(x) = \sum_{n=1}^\infty \frac{(-1)^{n-1} (2n+1)x^{2n}}{n^3}.$

Step 1: Truncate the series.

We approximate $g(x)$ by considering the first few terms. For $n = 1, 2, 3$: $g(x) \approx \frac{(-1)^0 (2 \cdot 1 + 1)x^{2 \cdot 1}}{1^3} + \frac{(-1)^1 (2 \cdot 2 + 1)x^{2 \cdot 2}}{2^3} + \frac{(-1)^2 (2 \cdot 3 + 1)x^{2 \cdot 3}}{3^3}.$ This simplifies to: $g(x) \approx 3x^2 - \frac{5x^4}{8} + \frac{7x^6}{27}.$

Step 2: Integrate term by term.

[ \int_0^1 g(x) , dx \approx \int_0^1 \left( 3x^2 - \frac{5x^4}{8} + \frac{7x^6}{27