(a) Since \(x>0\),

\( f(x)=3x-6>-6 \).

Answer: the range is \( f(x)>-6 \).

(b) Let \(y=g(x)\).

\( y=\frac4{(ax-3)^2} \).

Since \(x>\frac3a\), we have \( ax-3>0 \).

Therefore

\( ax-3=\frac2{\sqrt y} \).

\( x=\frac{3+\frac2{\sqrt y}}{a} \).

Hence

Answer: \( g^{-1}(x)=\frac{3+\frac2{\sqrt x}}{a} \), for \(x>0\).

(c) If \(a=2\), then

\( g^{-1}(x)=\frac{3+\frac2{\sqrt x}}{2} \).

Solve \(g^{-1}(x)=4\):

\( \frac{3+\frac2{\sqrt x}}{2}=4 \).

\( 3+\frac2{\sqrt x}=8 \).

\( \frac2{\sqrt x}=5 \).

\( \sqrt x=\frac25 \).

Answer: \( x=\frac4{25} \).

(d) \(fg(8)=6\) means \( f(g(8))=6 \).

Since \(f(x)=3x-6\),

\( 3g(8)-6=6 \).

\( g(8)=4 \).

Now

\( g(8)=\frac4{(8a-3)^2} \).

So

\( \frac4{(8a-3)^2}=4 \).

\( (8a-3)^2=1 \).

\( 8a-3=1 \) or \( 8a-3=-1 \).

So \( a=\frac12 \) or \( a=\frac14 \).

But \(g(8)\) is defined only if \(8>\frac3a\), which means \(a>\frac38\).

Therefore \( a=\frac14 \) is rejected.

Answer: \( a=\frac12 \).