Suppose you have a Normal distribution $X \sim N(\mu ,\sigma^2)$ we can transform the random variable into a standard Normal distribution using $Z=\frac{X-\mu}{\sigma}$. This is done so that we have $Z \sim N(0,1)$ which is easier to work with.
The cumulative distribution function of the standard Normal distribution is given by: $$\Phi (z)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^ze^{-(1/2)t^2}dt$$
Given $z>0$ and using symmetry ($p(Z\le 0)=0.5$) this is the same as: $$\Phi (z)=\frac{1}{2}+\frac{1}{\sqrt{2 \pi}}\int_{0}^ze^{-(1/2)t^2}dt$$
As mentioned in the comments the integral isn't nice but one method would be to use integration by parts. This will result in an approximation but that may be all that's required. I will perform a few calculations to illustrate: $$\int_{0}^ze^{-(1/2)t^2}dt=\left[te^{-(1/2)t^2}\right]_0^z-\int_0^z -t^2e^{-(1/2)t^2}dt\\\int_{0}^ze^{-(1/2)t^2}dt=\left[te^{-(1/2)t^2}\right]_0^z+\int_0^z t^2e^{-(1/2)t^2}dt\\ \int_{0}^ze^{-(1/2)t^2}dt=ze^{-(1/2)z^2}+\int_0^z t^2e^{-(1/2)t^2}dt$$
Now keep repeating the integration by parts process as far as necessary. The next result is $$\int_{0}^ze^{-(1/2)t^2}dt=ze^{-(1/2)z^2}+\frac{z^3}{3}e^{-(1/2)t^2}+\int_0^z \frac{t^4}{3}e^{-(1/2)t^2}dt$$
Having done the integration a few times a pattern emerges which we may utilise $$\int_{0}^ze^{-(1/2)t^2}dt=ze^{-(1/2)z^2}+\frac{z^3}{3}e^{-(1/2)t^2}+\frac{z^5}{15}e^{-(1/2)t^2}+...$$
We may tidy this up by factoring$$\int_{0}^ze^{-(1/2)t^2}dt=e^{-(1/2)z^2}\left(z+\frac{z^3}{3}+\frac{z^5}{15}+...\right)$$
Putting this altogether we have a way to compute the cdf for $z>0$ namely$$\Phi (z)=\frac{1}{2}+\frac{1}{\sqrt{2 \pi}}e^{-(1/2)z^2}\left(z+\frac{z^3}{3}+\frac{z^5}{15}+...\right)$$
Finally if we want the probability that $z$ is within a standard deviation $1$ of the mean $(0)$ we compute $2(\Phi(1)-0.5)$ This leads to $$p(-1<Z<1)\approx 2\times 0.338759=0.677581$$
We have established the fact that around $\frac{2}{3}$ of the distribution lies within a standard deviation of the mean. The other results you mention could likewise be formulated.
Note I don't claim integrating by parts is the most efficient numerical approximation there will be better more sophisticated methods.
Hope this helps.