Answer
$a.$ The slope is $m_{tan}=8$
$b.$ The equation is $y=8x+12$
$c.$ The graph is on the figure below.
Work Step by Step
$a.$ For the slope of the tangent line at point $P(a,f(a))$ we have by definition (1) with $a=-1$ and $f(a) = 4$:
$$m_{tan} = \lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}= \lim_{x\to-1}\frac{\frac{4}{x^2}-4}{x+1} = \lim_{x\to-1}\frac{\frac{4-4x^2}{x^2}}{x+1} = \lim_{x\to-1}\frac{4-4x^2}{x^2(x+1)}= \lim_{x\to-1}\frac{4(1-x^2)}{x^2(x+1)}=\lim_{x\to-1}\frac{4(1-x)(x+1)}{x^2(x+1)}=\lim_{x\to-1}\frac{4(1-x)}{x^2}=\frac{4(1-(-1))}{(-1)^2} =8.$$
$b.$ From definition (1) we have $y-f(a) = m_{tan} (x-a)$. Using this formula with $a=-1$, $f(a)=4$ and $m_{tan}=8$:
$$y-4=8(x-(-1))\Rightarrow y-4=8x+8$$
which gives
$$y=8x+12.$$
$c.$ The graph is on the figure below. The function is solid and the tangent is dashed.