Answer: (a) \(x=3\). (b) \((a,b)=(2,7)\) or \((0,3)\).

First rearrange the equation into standard quadratic form in \(x\):

\(4x^2-k=4kx-2\)

\(4x^2-4kx+(2-k)=0.\)

For this quadratic to have no real roots, its discriminant must be negative.

Here \(a=4\), \(b=-4k\), and \(c=2-k\), so

\(\Delta=b^2-4ac=(-4k)^2-4(4)(2-k).\)

Simplifying,

\(\Delta=16k^2-16(2-k)=16(k^2+k-2).\)

Factorise the quadratic expression:

\(k^2+k-2=(k+2)(k-1).\)

So the condition for no real roots is

\(16(k+2)(k-1)\lt 0.\)

The factor \(16\) is positive, so it does not affect the inequality. The product \((k+2)(k-1)\) is negative between its two roots, \(-2\) and \(1\).

Therefore

\(-2\lt k\lt 1.\)