Given the inequality \(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4), we can solve for x as follows:

\(\frac{1}{3}\)(x + 1) - 1 > \(\frac{1}{5}\)(x + 4)
=> \(\frac{5}{15}\)(x + 1) - \(\frac{15}{15}\) > \(\frac{3}{15}\)(x + 4)
=> \(\frac{5}{15}\)x + \(\frac{5}{15}\) - 1 > \(\frac{3}{15}\)x + \(\frac{12}{15}\)
=> \(\frac{2}{15}\)x > \(\frac{22}{15}\)
=> x > 11

Hence, the range of values of x which satisfies the given inequality is x > 11.