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 \(b = 8\).

Since \((x + 2)\) is a factor of \(p(x)\), we have \(p(-2) = 0\).

Substitute \(x = -2\) into \(p(x)\):

\(a(-2)^3 + 5(-2)^2 - 4(-2) + b = 0\)

\(-8a + 20 + 8 + b = 0\)

\(-8a + b + 28 = 0\)

\(-8a + b = -28\) \(\text{(Equation 1)}\)

Since the remainder when \(p(x)\) is divided by \((x + 1)\) is 2, we have \(p(-1) = 2\).

Substitute \(x = -1\) into \(p(x)\):

\(a(-1)^3 + 5(-1)^2 - 4(-1) + b = 2\)

\(-a + 5 + 4 + b = 2\)

\(-a + b + 9 = 2\)

\(-a + b = -7\) \(\text{(Equation 2)}\)

Subtract Equation 2 from Equation 1:

\((-8a + b) - (-a + b) = -28 - (-7)\)

\(-8a + a = -28 + 7\)

\(-7a = -21\)

\(a = 3\)

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

\(-3 + b = -7\)

\(b = -4\)

Thus, \(a = 3\) and \(b = -4\).

To divide \(6x^4 + x^3 - x^2 + 5x - 6\) by \(2x^2 - x + 1\), we perform polynomial long division.

1. Divide the leading term \(6x^4\) by \(2x^2\) to get \(3x^2\).

2. Multiply \(3x^2\) by \(2x^2 - x + 1\) to get \(6x^4 - 3x^3 + 3x^2\).

3. Subtract \(6x^4 - 3x^3 + 3x^2\) from \(6x^4 + x^3 - x^2 + 5x - 6\) to get \(4x^3 - 4x^2 + 5x - 6\).

4. Divide \(4x^3\) by \(2x^2\) to get \(2x\).

5. Multiply \(2x\) by \(2x^2 - x + 1\) to get \(4x^3 - 2x^2 + 2x\).

6. Subtract \(4x^3 - 2x^2 + 2x\) from \(4x^3 - 4x^2 + 5x - 6\) to get \(-2x^2 + 3x - 6\).

7. Divide \(-2x^2\) by \(2x^2\) to get \(-1\).

8. Multiply \(-1\) by \(2x^2 - x + 1\) to get \(-2x^2 + x - 1\).

9. Subtract \(-2x^2 + x - 1\) from \(-2x^2 + 3x - 6\) to get \(2x - 5\).

The quotient is \(3x^2 + 2x - 1\) and the remainder is \(2x - 5\).

Since \((2x + 1)\) is a factor of \(p(x)\), substituting \(x = -\frac{1}{2}\) into \(p(x)\) gives:

\(-\frac{6}{8} + \frac{1}{4}a - \frac{1}{2}b - 2 = 0\)

Simplifying, we get:

\(-\frac{3}{4} + \frac{1}{4}a - \frac{1}{2}b - 2 = 0\)

\(\frac{1}{4}a - \frac{1}{2}b = \frac{11}{4}\)

Multiplying through by 4 gives:

\(a - 2b = 11\) (Equation 1)

Since the remainder when \(p(x)\) is divided by \((x + 2)\) is \(-24\), substituting \(x = -2\) into \(p(x)\) gives:

\(-48 + 4a - 2b - 2 = -24\)

Simplifying, we get:

\(-50 + 4a - 2b = -24\)

\(4a - 2b = 26\)

Dividing through by 2 gives:

\(2a - b = 13\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(a = 11 + 2b\)

Substitute into Equation 2:

\(2(11 + 2b) - b = 13\)

\(22 + 4b - b = 13\)

\(3b = -9\)

\(b = -3\)

Substitute \(b = -3\) back into Equation 1:

\(a - 2(-3) = 11\)

\(a + 6 = 11\)

\(a = 5\)

Thus, \(a = 5\) and \(b = -3\).

We start by dividing \(p(x) = x^4 + 3x^3 + ax + b\) by \(x^2 + x - 1\). The division gives a quotient of the form \(x^2 + kx\) and a remainder \((a+3)x + (b-1)\).

Since the remainder is given as \(2x + 3\), we equate the coefficients:

\(a + 3 = 2\) and \(b - 1 = 3\).

Solving these equations gives:

\(a + 3 = 2 \Rightarrow a = -1\)

\(b - 1 = 3 \Rightarrow b = 4\)

Thus, the values are \(a = -1\) and \(b = 4\).