Answer
$a.$ $(f\circ g)(x)=-5x^{2}+20x-7$
$b.$ $(g\circ f)(x)=-25x^{2}+40x-13$
$c.$ $(f\circ g)(2)=13$
$d.$ $(g\circ f)(2)=-33$
Work Step by Step
$f(x)=5x-2$ $,$ $g(x)=-x^{2}+4x-1$
$a.$ $(f\circ g)(x)$
To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify:
$(f\circ g)(x)=f(g(x))=5(-x^{2}+4x-1)-2=...$
$...=-5x^{2}+20x-5-2=-5x^{2}+20x-7$
$b.$ $(g\circ f)(x)$
To find $(g\circ f)(x)$, substitute $x$ by $f(x)$ in $g(x)$ and simplify:
$(g\circ f)(x)=g(f(x))=-(5x-2)^{2}+4(5x-2)-1=...$
$...=-(25x^{2}-20x+4)+20x-8-1=...$
$...=-25x^{2}+20x-4+20x-8-1=-25x^{2}+40x-13$
$c.$ $(f\circ g)(2)$
Substitute $x$ by $2$ in $(f\circ g)(x)$, which was found in part $a$, and evaluate:
$(f\circ g)(2)=f(g(2))=-5(2)^{2}+20(2)-7=-5(4)+40-7=...$
$...=-20+40-7=13$
$d.$ $(g\circ f)(2)$
Substitute $x$ by $2$ in $(g\circ f)(x)$, which was found in part $b$, and evaluate:
$(g\circ f)(2)=g(f(2))=-25(2)^{2}+40(2)-13=...$
$...=-25(4)+80-13=-100+80-13=-33$