(a) The sum to infinity of the geometric progression is given by \(\frac{5a}{1 - \left(-\frac{1}{4}\right)}\).

The sum of the first eight terms of the arithmetic progression is \(\frac{8}{2} [2a + 7(-4)]\).

Equating the two sums: \(\frac{5a}{1 + \frac{1}{4}} = 4a + 8(-4)\).

Simplifying, \(\frac{5a}{\frac{5}{4}} = 8a - 28\).

\(4a = 8a - 112\).

Solving for \(a\), we get \(a = 28\).

(b) The \(k\)th term of the arithmetic progression is given by \(a + (k-1)(-4) = 0\).

Substituting \(a = 28\), we have \(28 + (k-1)(-4) = 0\).

Solving for \(k\), \(28 - 4k + 4 = 0\).

\(32 = 4k\).

\(k = 8\).