To find the values of \(a\) and \(b\), we use the Remainder Theorem. For \(p(x)\) divided by \((x + 2)\), substitute \(x = -2\) into \(p(x)\):

\(2(-2)^3 + a(-2)^2 + b(-2) + 6 = -38\)

\(-16 + 4a - 2b + 6 = -38\)

\(4a - 2b - 10 = -38\)

\(4a - 2b = -28\)

\(2a - b = -14\) \(\text{(Equation 1)}\)

For \(p(x)\) divided by \((2x - 1)\), substitute \(x = \frac{1}{2}\) into \(p(x)\):

\(2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 6 = \frac{19}{2}\)

\(\frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 6 = \frac{19}{2}\)

Multiply through by 4 to clear fractions:

\(1 + a + 2b + 24 = 38\)

\(a + 2b = 13\) \(\text{(Equation 2)}\)

Now solve the system of equations:

Equation 1: \(2a - b = -14\)

Equation 2: \(a + 2b = 13\)

Multiply Equation 1 by 2:

\(4a - 2b = -28\)

Add to Equation 2:

\(4a - 2b + a + 2b = -28 + 13\)

\(5a = -15\)

\(a = -3\)

Substitute \(a = -3\) into Equation 2:

\(-3 + 2b = 13\)

\(2b = 16\)

\(b = 8\)

Thus, \(a = -3\) and \(b = 8\).