#### Answer

$a.$ The value of the derivative is $f'(-3)=8.$
$b.$ The equation is $y=8x$.

#### Work Step by Step

$a.$ By definition of a derivative at point $a=-3$ we have
$$f'(-3)=\lim_{h\to0}\frac{f(-3+h)-f(-3)}{h}=\lim_{h\to0}\frac{8(-3+h)-8\cdot(-3)}{h}=\lim_{h\to0}\frac{-24+8h+24}{h}=\lim_{h\to0}\frac{8h}{h}=\lim_{h\to0}8=8.$$
$b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=-3$, $f(a)=8\cdot(-3)=-24$ and $f'(a)=8$ we get
$$y-(-24)=8(x-(-3))\Rightarrow y+24=8x+24$$ which gives $$y=8x.$$