Answer: \(k=-4\); term independent of \(y\) is \(7920\).

Use the binomial theorem and identify the required term carefully from the general term. Keep the powers of the variable explicit so that the required coefficient or constant term can be read off clearly.

In \((1+kx)^{15}\), the coefficient of \(x^3\) is

\(\binom{15}{3}k^3\).

So

\(\binom{15}{3}k^3=-29120\).

Since \(\binom{15}{3}=455\),

\(455k^3=-29120\), so \(k^3=-64\), and \(k=-4\).

For \(\left(8y^2-\frac{1}{2y}\right)^{12}\), the general term is

\(\binom{12}{r}(8y^2)^{12-r}\left(-\frac{1}{2y}\right)^r\).

The power of \(y\) is \(2(12-r)-r=24-3r\).

For the term independent of \(y\), \(24-3r=0\), so \(r=8\).

The required term is

\(\binom{12}{8}(8y^2)^4\left(-\frac{1}{2y}\right)^8=495\cdot8^4\cdot\frac{1}{2^8}=7920\).

This gives the required answer: \(k=-4\); term independent of \(y\) is \(7920\).