(a) The square is nonnegative and attains zero at \(x=2\), so \[\boxed{\text{range}=[-4,\infty)}.\]

(b) From \(y=(x-2)^2-4\), \(x-2=\sqrt{y+4}\) because \(x\ge2\). Thus \[\boxed{f^{-1}(x)=2+\sqrt{x+4},\quad x\ge-4}.\]

(c) \[x^2-4x=-\frac53x+2\quad\Longrightarrow\quad3x^2-7x-6=(3x+2)(x-3)=0.\] The root \(-2/3\) is outside the domain of f; hence \[\boxed{x=3}.\]

(d) \(f^{-1}(12)=2+\sqrt{16}=6\). Applying g twice gives \[g(6)=6a+2,\qquad g(g(6))=a(6a+2)+2=62.\] Thus \[6a^2+2a-60=2(3a+10)(a-3)=0,\] so \[\boxed{a=3\quad\text{or}\quad a=-\frac{10}{3}}.\]