To find the coefficient of \(x^3\) in the expansion of \((a+x)^5 + (1-2x)^6\), we need to consider the individual expansions.

For \((a+x)^5\), the term containing \(x^3\) is given by the binomial coefficient:

\(\binom{5}{3} a^2 x^3 = 10a^2 x^3\)

For \((1-2x)^6\), the term containing \(x^3\) is:

\(\binom{6}{3} (1)^3 (-2x)^3 = 20(-8)x^3 = -160x^3\)

The total coefficient of \(x^3\) in the expansion is:

\(10a^2 - 160 = 90\)

Solving for \(a\):

\(10a^2 = 250\)

\(a^2 = 25\)

\(a = 5\)