(i) Since \((x - 2)\) is a factor of \(p(x)\), substituting \(x = 2\) into \(p(x)\) should yield zero: \(2(2)^3 + a(2)^2 - 4 = 0\). Simplifying gives \(16 + 4a - 4 = 0\), leading to \(4a = -12\), so \(a = -3\).

(ii) With \(a = -3\), \(p(x) = 2x^3 - 3x^2 - 4\). Dividing by \((x - 2)\) gives the quadratic factor \(2x^2 + x + 2\). Thus, \(p(x) = (x-2)(2x^2 + x + 2)\).

(iii) To solve \(p(x) > 0\), note that \(2x^2 + x + 2\) has no real roots and is always positive. Therefore, \(p(x) > 0\) when \(x > 2\).

(i) Since \(f(x)\) is divisible by \(x^2 - 4x + 4\), we can write \(f(x) = (x^2 - 4x + 4)(x^2 + bx + c)\).

By comparing coefficients, we find that \(b = 2\) and \(4c = a\). Solving for \(a\), we get \(a = 8\).

(ii) Substitute \(a = 8\) into \(f(x)\), giving \(f(x) = (x^2 - 4x + 4)(x^2 + 2x + 2)\).

Since \(x^2 - 4x + 4 = (x-2)^2\), it is always non-negative.

For \(x^2 + 2x + 2\), the discriminant is \(2^2 - 4 \times 1 \times 2 = -4\), which is negative, indicating no real roots and always positive.

Thus, \(f(x)\) is the product of two non-negative expressions, making \(f(x)\) never negative.

Given that \((x^2 + x + 2)\) is a factor of \(p(x) = x^4 + 4x^2 + x + a\), we can express \(p(x)\) as:

\(p(x) = (x^2 + x + 2)(x^2 + bx + c)\)

Expanding the right-hand side, we have:

\((x^2 + x + 2)(x^2 + bx + c) = x^4 + bx^3 + cx^2 + x^3 + bx^2 + cx + 2x^2 + 2bx + 2c\)

Combining like terms, we get:

\(x^4 + (b+1)x^3 + (c+b+2)x^2 + (c+2b)x + 2c\)

Equating coefficients with \(x^4 + 4x^2 + x + a\), we have:

1. \(b+1 = 0\) implies \(b = -1\)

2. \(c+b+2 = 4\) implies \(c - 1 + 2 = 4\) so \(c = 3\)

3. \(c+2b = 1\) implies \(3 - 2 = 1\), which is consistent

4. \(2c = a\) implies \(2 \times 3 = a\) so \(a = 6\)

Thus, the other quadratic factor is \(x^2 - x + 3\).

To find the values of \(a\) and \(b\), we perform polynomial division of \(2x^4 + ax^3 + bx - 1\) by \(x^2 - x + 1\) and set the remainder equal to \(3x + 2\).

1. Start the division: The first term of the quotient is \(2x^2\) because \(2x^2 \cdot (x^2 - x + 1) = 2x^4 - 2x^3 + 2x^2\).

2. Subtract \(2x^4 - 2x^3 + 2x^2\) from \(2x^4 + ax^3 + bx - 1\) to get \((a + 2)x^3 - 2x^2 + bx - 1\).

3. The next term of the quotient is \((a + 2)x\) because \((a + 2)x \cdot (x^2 - x + 1) = (a + 2)x^3 - (a + 2)x^2 + (a + 2)x\).

4. Subtract \((a + 2)x^3 - (a + 2)x^2 + (a + 2)x\) from \((a + 2)x^3 - 2x^2 + bx - 1\) to get \((b - (a + 2))x - 1\).

5. The remainder is \((b - (a + 2))x - 1\), which must equal \(3x + 2\).

6. Equate coefficients: \(b - (a + 2) = 3\) and \(-1 = 2\).

7. Solve the equations: From \(-1 = 2\), we find \(a = -3\). From \(b - (a + 2) = 3\), substitute \(a = -3\) to get \(b - (-3 + 2) = 3\), which simplifies to \(b + 1 = 3\), so \(b = 5\).

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

Since \(p(x)\) is divisible by \((2x - 1)\), we have \(p\left(\frac{1}{2}\right) = 0\).

Substitute \(x = \frac{1}{2}\) into \(p(x)\):

\(a\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 3 = 0\)

\(\frac{1}{8}a + \frac{1}{4} + \frac{1}{2}b + 3 = 0\)

Multiply through by 8 to clear fractions:

\(a + 2 + 4b + 24 = 0\)

\(a + 4b = -26\)

When \(p(x)\) is divided by \((x + 2)\), the remainder is 5, so \(p(-2) = 5\).

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

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

\(-8a + 4 - 2b + 3 = 5\)

\(-8a - 2b = -2\)

\(8a + 2b = 2\)

Now solve the system of equations:

1. \(a + 4b = -26\)

2. \(8a + 2b = 2\)

Multiply equation 1 by 2:

\(2a + 8b = -52\)

Subtract equation 2 from this result:

\(2a + 8b - 8a - 2b = -52 - 2\)

\(-6a + 6b = -54\)

\(-a + b = -9\)

\(b = a - 9\)

Substitute \(b = a - 9\) into equation 1:

\(a + 4(a - 9) = -26\)

\(a + 4a - 36 = -26\)

\(5a = 10\)

\(a = 2\)

Substitute \(a = 2\) back to find \(b\):

\(b = 2 - 9 = -7\)

Thus, \(a = 2\) and \(b = -7\).