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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$\ds{I \equiv \int_{0}^{\infty}x^{2}\expo{-x^{2}}{\rm erf}\pars{x}\ln\pars{x}
\,\dd x}$. Let's
$\ds{{\cal I}\pars{\mu}
\equiv \int_{0}^{\infty}x^{\mu}\expo{-x^{2}}{\rm erf}\pars{x}\,\dd x}$ such that $\ds{I = \lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}}$
Since
$\ds{{\rm erf}\pars{x}
\stackrel{{\rm def.}}{=}{2 \over \root{\pi}}\int_{0}^{x}\expo{-y^{2}}\,\dd y}$:
\begin{align}
{\cal I}\pars{\mu}&=\int_{0}^{\infty}x^{\mu}\expo{-x^{2}}
{2 \over \root{\pi}}\int_{0}^{\infty}\Theta\pars{x - y}\expo{-y^{2}}\,\dd y\,\dd x
\\[3mm]&=
{2 \over \root{\pi}}\int_{0}^{\pi/2}\dd\theta\,\cos^{\mu}\pars{\theta}
\Theta\pars{\cos\pars{\theta} - \sin\pars{\theta}}
\overbrace{\int_{0}^{\infty}\dd r\,r^{\mu + 1}\expo{-r^{2}}}
^{\Gamma\pars{1 + \mu/2}/2}
\\[3mm]&={1 \over \root{\pi}}\,\Gamma\pars{1 + {\mu \over 2}}
\int_{0}^{\pi/4}\dd\theta\,\cos^{\mu}\pars{\theta}
\end{align}
\begin{align}
I&=\lim_{\mu \to 2}\totald{{\cal I}\pars{\mu}}{\mu}=
{1 \over 2\root{\pi}}\,\overbrace{\Psi\pars{2}}^{\ds{1 - \gamma}}\
\overbrace{\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}}^{\ds{\pars{\pi + 2}/8}}
\\[3mm]&+
{1 \over \root{\pi}}\,\overbrace{\Gamma\pars{2}}^{\ds{1}}
\overbrace{%
\int_{0}^{\pi/4}\dd\theta\,\cos^{2}\pars{\theta}\ln\pars{\cos\pars{\theta}}}
^{\ds{\braces{4G + \pi - 2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}}/16}}
\end{align}
$\Gamma\pars{z}$ and $\Psi\pars{z}$ are the $\it Gamma$ and $\it Digamma$ functions, respectively. $\gamma$ and $G$ are the $\it Euler-Mascheroni$ and Catalan constants, respectively.
$$
\begin{array}{|l|}\hline \mbox{}\\
\quad{\displaystyle\int_{0}^{\infty}x^{2}\expo{-x^{2}}
\,\mathrm{erf}\pars{x}\ln\pars{x}\,\dd x}\quad
\\[2mm] =
\quad{{\displaystyle\quad\pars{\pi + 2}\pars{1 - \gamma} + 4G + \pi -
2\bracks{1 + \ln\pars{2}} - \pi\ln\pars{4}\quad} \over
{\displaystyle 16\root{\pi}}}\quad
\\ \mbox{}\\ \hline
\end{array}
\approx 0.0436462
$$
Best Answer
By De Moivre's formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ we have the following Fourier sine series: $$\frac{\sin(3x)\sin(4x)\sin(5x)\cos(6x)}{\sin^2(x)}\\= -\frac{1}{2} \sin(2x)-\frac{1}{2}\sin(4x)+\sin(8x)+\frac{3}{2}\sin(10x)+\frac{3}{2}\sin(12x)+\sin(14x)+\frac{1}{2}\sin(16 x)$$ and: $$I(n)=\int_{0}^{+\infty}\frac{\sin(2nx)}{x\cosh(x)}\,dx = 2\arctan\left(\tanh\frac{\pi n}{2}\right) $$ follows by differentiation under the integral sign. The original integral can so be expressed in terms of the Gudermannian function:
$$ I = \frac{1}{2} \big(-\text{gd}(\pi)- \text{gd}(2\pi) + 2 \text{gd}(4\pi) + 3 \text{gd}(5\pi) + 3 \text{gd}(6\pi) + 2 \text{gd}(7\pi) + \text{gd}(8\pi)\big) \approx 7.11363 $$