I can address the second integral:
$$\int_{0}^{\infty }{dx \: \frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}}$$
Hint: We can use Parseval's Theorem
$$\int_{-\infty}^{\infty} dx \: f(x) \bar{g}(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{f}(k) \bar{\hat{g}}(k) $$
where $f$ and $\hat{f}$ are Fourier transform pairs, and same for $g$ and $\bar{g}$. The FT of $1/(x^2+\pi^2)$ is easy, so we need the FT of the rest of the integrand, which turns out to be possible.
Define
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x} $$
It is straightforward to show using the Residue Theorem that, when $f(x) = (x^2+a^2)^{-1}$, then
$$\hat{f}(k) = \frac{\pi}{a} e^{-a |k|} $$
Thus we need to compute, when $g(x) = (x-\sin{x})/x^3$,
$$\begin{align} \hat{g}(k) &= \int_{-\infty}^{\infty} dx \: \frac{x-\sin{x}}{x^3} e^{i k x} \\ &= \frac{\pi}{2}(k^2-2 |k|+1) \mathrm{rect}(k/2) \\ \end{align}$$
where
$$\mathrm{rect}(k) = \begin{cases} 1 & |k|<\frac{1}{2} \\ 0 & |k|>\frac{1}{2} \end{cases} $$
Then we can write, using the Parseval theorem,
$$\begin{align} \int_{0}^{\infty }{dx \: \frac{x-\sin x}{\left( {{\pi }^{2}}+{{x}^{2}} \right){{x}^{3}}}} &= \frac{1}{8} \int_{-1}^1 dk \: (k^2-2 |k|+1) e^{-\pi |k|} \\ &= \frac{\left(2-2 \pi +\pi ^2\right)}{4 \pi
^3}-\frac{ e^{-\pi }}{2 \pi ^3} \\ \end{align}$$
NOTE
Deriving $\hat{g}(k)$ from scratch is challenging; nevertheless, it is straightforward (albeit, a bit messy) to prove that the expression is correct by performing the inverse transform on $\hat{g}(k)$ to obtain $g(x)$. I did this out and proved it to myself; I can provide the details to those that want to see them.
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\begin{align}
&\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\,
\cos\pars{x \over 2}\,\dd x \\[5mm] = &
\int_{0}^{\infty}{x\bracks{1 - 2\sin^{2}\pars{x/2}} - \sin\pars{x} \over x^{3}}\,\cos\pars{x \over 2}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\infty}{2x\cos\pars{x/2} - 2x\sin\pars{x}\sin\pars{x/2} -
2\sin\pars{x}\cos\pars{x/2} \over x^{3}}\,\,\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{\infty}{x\cos\pars{x/2} + x\cos\pars{3x/2} -
\sin\pars{3x/2} - \sin\pars{x/2} \over x^{3}}\,\,\,\dd x
\\[1cm] & =
-\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -
{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & -
{1 \over 4}\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{2x - \sin\pars{3x/2} - \sin\pars{x/2}}\,\dd\pars{1 \over x^{2}}
\end{align}
Integrating by parts the last integral:
\begin{align}
&\int_{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x^{3}}\,
\cos\pars{x \over 2}\,\dd x =
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -
{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & +
{1 \over 4}\int_{x = 0}^{\infty}{2 - 3\cos\pars{3x/2}/2 - \cos\pars{x/2}/2 \over x^{2}}\,\dd x
\\[1cm] & =
-\,{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x -
{1 \over 2}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & +
{3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x +
{1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x
\\[1cm] & =
-\,{3 \over 8}\int_{0}^{\infty}{1 - \cos\pars{x/2} \over x^{2}}\,\dd x
-\,{1 \over 8}\int_{0}^{\infty}{1 - \cos\pars{3x/2} \over x^{2}}\,\dd x
\\[5mm] & =
-\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x
-\,{3 \over 16}\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x =
-\,{3 \over 8}\
\int_{0}^{\infty}{1 - \cos\pars{x} \over x^{2}}\,\dd x
\\[5mm] & =
-\,{3 \over 4}\int_{0}^{\infty}{\sin^{2}\pars{x/2} \over x^{2}}\,\dd x =
-\,{3 \over 8}\
\underbrace{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x}
_{\ds{=\ {\pi \over 2}}}\ = \
\bbox[#ffe,10px,border:1px dotted navy]{\ds{-\,{3 \over 16}\,\pi}}
\end{align}
>By integrating by parts:
$\ds{\int_{0}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,\dd x =
\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x = {1 \over 2}\,\pi}$.
Best Answer
Hint The form of the integral over $(-\infty, \infty)$ suggests using residue calculus. Here's an outline; I'll let you fill in the details.
Remark Instead taking the imaginary part of the quantity in brackets gives the value of a related integral: $$\int_0^\infty \frac{\cos t^2 + \sin t^2}{1 + t^4} \,dt = \frac{\pi e}{2 \sqrt 2} .$$