Identify coefficients and Vieta's formulas for the given quadratic equation.

Step 1: Identify coefficients and Vieta's formulas for the given quadratic equation.
The given quadratic equation is $3x^2 - (p-4)x + (2p+1) = 0$.
Comparing this to $Ax^2 + Bx + C = 0$, we have:
$A = 3$
$B = -(p-4) = 4-p$
$C = 2p+1$

The sum of the roots ($\alpha + \beta$) and the product of the roots ($\alpha \beta$) are given by Vieta's formulas:

$$\alpha + \beta = -\frac{B}{A} = -\frac{-(p-4)}{3} = \frac{p-4}{3}$$ $$\alpha \beta = \frac{C}{A} = \frac{2p+1}{3}$$

Step 2: Use the given condition to find the value of $p$.
The problem states that $(\alpha+2)(\beta+2) = 1$.
Expand this expression:

$$\alpha\beta + 2\alpha + 2\beta + 4 = 1$$ $$\alpha\beta + 2(\alpha + \beta) + 4 = 1$$

Substitute the expressions for $\alpha + \beta$ and $\alpha \beta$ from Step 1:

$$\frac{2p+1}{3} + 2\left(\frac{p-4}{3}\right) + 4 = 1$$

Multiply the entire equation by 3 to eliminate the denominators:

$$3\left(\frac{2p+1}{3}\right) + 3\left(\frac{2(p-4)}{3}\right) + 3(4) = 3(1)$$ $$(2p+1) + 2(p-4) + 12 = 3$$ $$2p+1 + 2p-8 + 12 = 3$$

Combine like terms:

$$4p + 5 = 3$$ $$4p = 3 - 5$$ $$4p = -2$$ $$p = -\frac{2}{4}$$ $$p = -\frac{1}{2}$$