= Solution
Evaluate the <Fourier sine series> of part (d) at the interior <point> $x=1/2$. Only odd indices remain, with alternating signs:
$$
\frac12=\frac2\pi\left(1-\frac13+\frac15-\cdots\right),\qquad \pi=4\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}.
$$
Thus $\boxed{A=4}$.
For the second identity, apply <Parseval's identity> to $h(x)=x^2$. From part (c), $a_0=2/3$, $a_n=4(-1)^n/(n^2\pi^2)$ and $b_n=0$, so
$$
\int_{-1}^1x^4\,dx=\frac12\left(\frac23\right)^2+\frac{16}{\pi^4}\sum_{n=1}^{\infty}\frac1{n^4}.
$$
The left side is $2/5$, giving $\frac{16}{\pi^4}\sum n^{-4}=2/5-2/9=8/45$. Consequently
$$
\boxed{\pi^4=90\sum_{n=1}^{\infty}\frac1{n^4},\qquad B=90.}
$$
The first <series> is conditionally convergent, while the second converges absolutely by the <P-series> criterion.
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