(a) \[y^{\prime}=-\frac{24}{(3x-8)^2}+\frac6{(x-1)^2}.\] At \(x=3\), the gradient is \(-45/2\). The tangent is \(y-5=-(45/2)(x-3)\). With \(y=-8x\), \[-16x-10=-45x+135\Rightarrow x=5,y=-40.\]

(b)(i) \[\frac6{(x-1)^2}=\frac{24}{(3x-8)^2}\Rightarrow(3x-8)^2=4(x-1)^2\Rightarrow5(x-2)(x-6)=0.\] Hence \(x=2,6\), both in the curve’s domain.

(b)(ii) \[y^{\prime\prime}=\frac{144}{(3x-8)^3}-\frac{12}{(x-1)^3}.\] At \(x=2\), \(y^{\prime\prime}=-30<0\): maximum. At \(x=6\), \(y^{\prime\prime}=6/125>0\): minimum.