Answer
$a.$ The slope is $m_{tan}=4.$
$b.$ The equation is $y=4x-8.$
Work Step by Step
$a.$ Using the formula from definition (2) with $a=2$ and $f(a)=0$ (coordinates of the point $P(2,0)$) we have
$$m_{tan}=\lim_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0}\frac{(2+h)^2-4-0}{h}=\lim_{h\to0}\frac{4+h^2+4h-4}{h}=\lim_{h\to0}\frac{h^2+4h}{h}=\lim_{h\to0}(h+4)=0+4=4.$$
$b.$ Using the formula $y-f(a)=m_{tan}(x-a)$ with the same values for $a$ and $f(a)$ as in part $a$ and the calculated value $m_{tan} = 4$ we get
$$y-0=4(x-2)\Rightarrow y=4x-8.$$
