Answer: \((x,y)=(4,8)\) or \((-2,2)\).

From

\(y-x=4,\)

we have

\(y=x+4.\)

Substitute this into the second equation:

\(x^2+(x+4)^2-8x-4(x+4)-16=0.\)

Expand and simplify:

\(x^2+x^2+8x+16-8x-4x-16-16=0.\)

So

\(2x^2-4x-16=0.\)

Divide by \(2\):

\(x^2-2x-8=0.\)

Factorise:

\((x-4)(x+2)=0.\)

Hence

\(x=4\quad\text{or}\quad x=-2.\)

If \(x=4\), then \(y=4+4=8\). If \(x=-2\), then \(y=-2+4=2\). Therefore

\(\boxed{(x,y)=(4,8)\quad\text{or}\quad(-2,2)}.\)