First, consider the inequality \(|2^{x+1} - 2| < 0.5\). This can be rewritten as two separate inequalities:

1. \(2^{x+1} - 2 < 0.5\)

2. \(2^{x+1} - 2 > -0.5\)

For the first inequality:

\(2^{x+1} < 2.5\)

Taking the logarithm base 2, we have:

\((x+1)\ln 2 < \ln 2.5\)

\(x < \frac{\ln 2.5}{\ln 2} - 1\)

For the second inequality:

\(2^{x+1} > 1.5\)

Taking the logarithm base 2, we have:

\((x+1)\ln 2 > \ln 1.5\)

\(x > \frac{\ln 1.5}{\ln 2} - 1\)

Calculating the critical values:

\(x < \frac{\ln 2.5}{\ln 2} - 1 \approx 0.322\)

\(x > \frac{\ln 1.5}{\ln 2} - 1 \approx -0.415\)

Thus, the solution is \(-0.415 < x < 0.322\).