Rational function approximation to square root

Concerning the first question $$\sqrt{x}\approx\frac{x^4+112 x^3+1120 x^2+1792 x+256}{16 x^3+448 x^2+1792 x+1024}$$ is the $\text{Pade}_{(4,3)}$ expansion of $\sqrt x$ centered at $x=4$ (have a look here).

Around a given point, any function can be approximated using Padé approximants which are ratios of polynomial of selected degrees.

If you consider $\text{Pade}_{(4,4)}$, it would be $$\sqrt{x}\approx \frac{9 x^4+336 x^3+2016 x^2+2304 x+256}{x^4+144 x^3+2016 x^2+5376 x+2304}$$ which is much better.

For the same number of coefficients, Padé approximants are incredibly more powerful than Taylor expansions. For example, in the case you gave, over the range $3\leq x \leq 5$, the largest error is less than $2.5 \times 10^{-9}$ that is to say more than ten times smaller than the result of a Taylor expansion to $O\left((x-4)^8\right)$.

We could play a lot with this kind of work and be sure that it is used a lot for many problems.

Edit (four years later)

Concerning the first part of the problem, we can make it more general and write (keeping the idea of a $[4,4]$ Padé approximant built at $x=a$) $$\sqrt[n] x=\sqrt[n] a\frac{1+b_1 t+b_2t^2+b_3t^3+b_4t^4}{1+c_1 t+c_2t^2+c_3t^3+c_4t^4} \quad \text{where} \quad t=\frac{x-a}{2 a n}$$ where the coefficients are $$b_1=4n+1$$ $$ b_2=\frac{3(3 n+1) (4 n+1)}{7}$$ $$b_3=\frac{2(2 n+1) (3 n+1) (4 n+1)}{21}$$ $$ b_4=\frac{(n+1) (2 n+1) (3 n+1) (4 n+1)}{105} $$ $$c_1=4n-1$$ $$ c_2=\frac{3(3 n-1) (4 n-1)}{7}$$ $$c_3=\frac{2(2 n-1) (3 n-1) (4 n-1)}{21}$$ $$ c_4=\frac{(n-1) (2 n-1) (3 n-1) (4 n-1)}{105} $$

It looks like the OP started with the observation that, when $(3+\sqrt8)^8$ is expanded into an expression of the form $a+b\sqrt8$, then $\sqrt8\approx a/b$, and similarly for other square roots. There is a fairly simple explanation for all this, based on examining the conjugate expression(s), $(3-\sqrt8)^8=(0.1715729\ldots)^8\approx0.00000075$.

If $(3+\sqrt8)^8=a+b\sqrt8$, then $(3-\sqrt8)^8=a-b\sqrt8$, hence $(3+\sqrt8)^8+(3-\sqrt8)^8=2a$. But since $3-\sqrt8=0.1715729\ldots$ is less than $1$, its eight power is much less than $1$, and so we have

$$2a=(3+\sqrt8)^8+(3-\sqrt8)^8\approx(3+\sqrt8)^8=a+b\sqrt8$$

from which $\sqrt8\approx a/b$ follows.