(a) Substitution gives \(\tfrac54x^2+(3-2a)x+a^2-11=0\). A repeated root requires \((3-2a)^2-5(a^2-11)=0\), simplifying to \((a-4)(a+16)=0\). Thus \(\boxed{a=4,-16}\), verifying the given value.

(b) The normal gradient is \(-2\), and it passes through the centre \((4,3)\). Thus \(\boxed{y=-2x+11}\). The contact point with \(y=x/2+6\) is \(P=(2,7)\).

(c) Parallel tangents have form \(y=-2x+c\). Their distance from \((4,3)\) is \(|11-c|/\sqrt5\), equal to radius \(\sqrt{20}=2\sqrt5\). Hence \(|11-c|=10\), so \(c=1,21\). Therefore \(\boxed{y=-2x+1,\ y=-2x+21}\).