(a) To solve \(\int_0^a xe^{-2x} \, dx = \frac{1}{8}\), we start by integrating by parts. Let \(u = x\) and \(dv = e^{-2x} \, dx\). Then \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\).

Using integration by parts, \(\int xe^{-2x} \, dx = -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} \, dx\).

Integrating \(\int e^{-2x} \, dx\) gives \(-\frac{1}{2}e^{-2x}\), so \(\int xe^{-2x} \, dx = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\).

Evaluating from 0 to a, we have:

\(\left[ -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \right]_0^a = -\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4}\).

Setting this equal to \(\frac{1}{8}\), we solve:

\(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4} = \frac{1}{8}\).

Rearranging gives \(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} = -\frac{1}{8}\).

Multiplying through by -1 and simplifying, we find \(a = \frac{1}{2} \ln(4a + 2)\).

(b) To verify a lies between 0.5 and 1, calculate:

For \(a = 0.5\), \(\frac{1}{2} \ln(4 \times 0.5 + 2) = \frac{1}{2} \ln(4) = 0.693...\).

For \(a = 1\), \(\frac{1}{2} \ln(4 \times 1 + 2) = \frac{1}{2} \ln(6) = 0.895...\).

Since \(0.5 < 0.693 < 1\), a is between 0.5 and 1.

(c) Using the iterative formula \(a_{n+1} = \frac{1}{2} \ln(4a_n + 2)\), start with \(a_0 = 0.75\):

\(a_1 = \frac{1}{2} \ln(4 \times 0.75 + 2) = 0.8047\)

\(a_2 = \frac{1}{2} \ln(4 \times 0.8047 + 2) = 0.8261\)

\(a_3 = \frac{1}{2} \ln(4 \times 0.8261 + 2) = 0.8343\)

\(a_4 = \frac{1}{2} \ln(4 \times 0.8343 + 2) = 0.8373\)

\(a_5 = \frac{1}{2} \ln(4 \times 0.8373 + 2) = 0.8385\)

Continuing this process, we find \(a \approx 0.84\) to 2 decimal places.