Answer
$a.$ The slope is $m_{tan} = 3$.
$b.$ The equation is $y=3x-2$.
Work Step by Step
$a.$ Using the formula from definition (2) with $a=1$ and $f(a)=1$ (coordinates of the point $P(1,1)$) we have
$$m_{tan}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{(1+h)^3-1}{h}=\lim_{h\to0}\frac{h^3+3h^2+3h+1-1}{h}=\lim_{h\to0}\frac{h^3+3h^2+3h}{h}=\lim_{h\to0}(h^2+3h+3)=0^2+3\cdot0+3=3.$$
$b.$ 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 $m_{tan} = 3$ we get
$$y-1=3(x-1)\Rightarrow y-1=3x-3$$ which gives
$$y=3x-2.$$
