Answer: (a) \(y+2=\frac45(x-8)\). (b) \(C=(13,2)\).

(a) The midpoint of \(AB\) is

\(\left(\frac{4+12}{2},\frac{3+(-7)}{2}\right)=(8,-2).\)

The gradient of \(AB\) is

\(\frac{-7-3}{12-4}=-\frac{10}{8}=-\frac54.\)

The gradient of the perpendicular bisector is therefore

\(\frac45.\)

Using the point \((8,-2)\), the equation of \(L\) is

\(\boxed{y+2=\frac45(x-8)}.\)

(b) The line through \((5,12)\) parallel to \(AB\) has gradient \(-\frac54\), so its equation is

\(y-12=-\frac54(x-5).\)

Write both lines in terms of \(y\):

\(L:\quad y=\frac45x-\frac{42}{5},\)

and

\(y=-\frac54x+\frac{73}{4}.\)

At \(C\), these are equal:

\(\frac45x-\frac{42}{5}=-\frac54x+\frac{73}{4}.\)

Multiplying by \(20\),

\(16x-168=-25x+365.\)

So

\(41x=533,\)

and hence \(x=13\). Substituting into \(L\),

\(y=\frac45(13)-\frac{42}{5}=2.\)

Therefore

\(\boxed{C=(13,2)}.\)