Answer
$a.\displaystyle \quad x_{n+1}=\frac{x_{n}+a/x_{n}}{2}$ verified below.
$b.\quad \sqrt{3}$ is being approximated. See below.
Work Step by Step
$ a.\quad$
$f(x)=x^{2}-a,\quad a\gt 0$
$f'(x)=2x$
$\begin{align*}
x_{n+1}& =x_{n}-\displaystyle \frac{x_{n}^{2}-a}{2x_{n}} \\
& =\displaystyle \frac{2x_{n}^{2}-(x_{n}^{2}-a)}{2x_{n}}\\
& =\displaystyle \frac{x_{n}^{2}+a}{2x_{n}} \\
&=\displaystyle \frac{x_{n}(x_{n}+a/x_{n})}{2x_{n}} \\
&=\displaystyle \frac{x_{n}+a/x_{n}}{2} \end{align*}.$
$ b.\quad$
$x_{0}=1, a=3$
$x_{1}=\displaystyle \frac{1+\frac{3}{1}}{2}=2$
$x_{2}=\displaystyle \frac{2+\frac{3}{2}}{2}=1.75$
$x_{3}=\displaystyle \frac{1.75+\frac{3}{1.75}}{2}=$1.73214285714286
Using a spreadsheet, if column B contains the values for $x_{n}$,
we place the formula =(B2+3/B2)/(2) into the cell B3, and copy-paste down.
n x
0 1
1 2
2 1.75
3 1.73214285714286
4 1.73205081001473
5 1.73205080756888
6 1.73205080756888
7 1.73205080756888
The number being approximated is the solution to the equation
$x^{2}-3=0, \quad x\gt 0$.
That is, x=$\sqrt{3}.$
