(i) To find \(a\), perform polynomial division of \(x^4 + 5x + a\) by \(x^2 - x + 3\). The quotient is \(x^2 + x\) and the remainder is \(3x + a\). Equating the remainder to zero gives:

\(3x + a = 0\)

Since \(x^2 - x + 3\) is a factor, the remainder must be zero for all \(x\). Thus, \(a = -6\).

Now, factorise \(x^2 + x - 2\) as \((x + 2)(x - 1)\). Therefore, \(p(x) = (x^2 - x + 3)(x + 2)(x - 1)\).

(ii) The quadratic \(x^2 + x - 2 = 0\) has two real roots, \(x = -2\) and \(x = 1\). The quadratic \(x^2 - x + 3 = 0\) has no real roots (discriminant \(b^2 - 4ac = 1 - 12 = -11\)). Therefore, \(p(x) = 0\) has two real roots.