Answer: (a) \(\cosh 2x=2\sinh^2 x+1\).

(b) The equation has two distinct real roots when \(k\lt -2\sqrt{2}\) or \(k\gt 2\sqrt{2}\).

(a) Using the definitions

\(\cosh x=\frac{1}{2}(e^x+e^{-x})\), \(\sinh x=\frac{1}{2}(e^x-e^{-x})\).

Then

\(2\sinh^2 x+1=2\left(\frac{1}{2}(e^x-e^{-x})\right)^2+1\)

\(=\frac{1}{2}(e^x-e^{-x})^2+1\)

\(=\frac{1}{2}(e^{2x}-2+e^{-2x})+1\)

\(=\frac{1}{2}(e^{2x}+e^{-2x})\)

But \(\cosh 2x=\frac{1}{2}(e^{2x}+e^{-2x})\), so \(\cosh 2x=2\sinh^2 x+1\).

(b) Substitute the result from part (a) into \(\cosh 2x=k\sinh x\):

\(2\sinh^2 x+1=k\sinh x\)

so

\(2\sinh^2 x-k\sinh x+1=0\).

Let \(y=\sinh x\). Since \(\sinh x\) can take any real value, this becomes

\(2y^2-ky+1=0\).

For two distinct real roots, the discriminant must be positive:

\((-k)^2-4(2)(1)\gt 0\)

\(k^2-8\gt 0\).

Hence \(k^2\gt 8\), so \(|k|\gt 2\sqrt{2}\).

Therefore the required values are \(k\lt -2\sqrt{2}\) or \(k\gt 2\sqrt{2}\).