(f-h)(x)=2x+5-7+x/3=7x/3-2, (f-h)(4)=28/3-2=22/3.
(f ○ g)(x)=f(g(x))=2g(x)+5=2(x2-3)+5=2x2-1.
(h ○ g)(x)=h(g(x))=7-g(x)/3=7-(x2-3)/3=7-x2/3+1=8-x2/3.
Let y=f(x)=2x+5, 2x=y-5, x=(y-5)/2=F(y), so F(x)=(x-5)/2. But F(x)=f-1(x), so f-1(x)=(x-5)/2.
(f ○ f-1)(x)=f(f-1(x))=2f-1(x)+5=2[(x-5)/2]+5=x (which is the definition of the inverse).
Similarly, let y=h(x)=7-x/3, 3y=21-x, x=21-3y=H(y), so h-1(x)=H(x)=21-3x.