$\quad\quad\quad$First question: $1 + \frac14\left(x^8 - 2 +\frac{1}{x^8}\right)\tag{1.1}$ $=\frac14\cdot 4 + \frac14\left(x^8 - 2 + +\frac{1}{x^8}\right)\quad=\quad\frac14\left(4 + x^8 -2 +\frac{1}{x^8}\right)\tag{1.2}$
$=\frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$
$=\frac14\left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$
$(1.3) \to (1.4):\quad\quad\text{Note that}$ $\frac14 \left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$ $= \frac14\left(x^{2\cdot 4} + 2\frac{x^4}{x^4} + \frac{1}{x^{2\cdot 4}}\right) = \frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$
For your second question: in your work you have:
$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]+1\tag{a}$
$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})+1]^2\tag{b}$
$1+[y']^2=[\frac{1}{2}(e^x-e^{-x}+2)]^2\tag{c}$
How did you get from $(a)\to(b)$? Did you forget the exponent in $(a)$? Shouldn't $(a)$ be $1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]^2+1\quad?$And if so, you cannot bring the constant term $1$ into an expression that is exponentiated without first making appropriate computations on the exponentiation expression. Failing that, your move from $(b)\to (c)$ is affected.
In this case, first expand $\left[\dfrac{1}{2}(e^x-e^{-x})\right]^2$, then worry about adding the $1$ term:
$1+[y']^2=\left[\frac{1}{2}(e^x-e^{-x})\right]^2+1\tag{2.1}$ $=\frac14\left(e^{2x} -2\frac{e^{2x}}{e^{2x}} + \frac{1}{e^{2x}}\right) + 1\tag{2.2}$
$=\frac14\left(e^{2x} -2 + \frac{1}{e^{2x}}\right) + \frac14\cdot4 \quad=\quad\frac14\left(e^{2x} -2 +\frac{1}{e^{2x}}+4\right)\tag{2.3}$ $= \left(\frac12\right)^2\left(e^{2x} +2 +\frac{1}{e^{2x}}\right)\quad=\quad\left[\frac12(e^x + e^{-x})\right]^2\tag{2.4}$