1. **State the problem:** We need to find the vertical asymptotes, horizontal asymptotes, and holes of the function $$f(x) = \frac{(x+1)(x+2)}{(x+2)(x-3)}.$$\n\n2. **Identify vertical asymptotes and holes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at that point. Holes occur where both numerator and denominator are zero at the same point (common factors).\n\n3. **Factor and simplify:** The numerator is $(x+1)(x+2)$ and the denominator is $(x+2)(x-3)$. The common factor is $(x+2)$.\n\n4. **Simplify the function:** Cancel the common factor $(x+2)$ to get $$f(x) = \frac{x+1}{x-3}, \quad x \neq -2.$$\n\n5. **Find holes:** The canceled factor $(x+2)$ corresponds to a hole at $x = -2$. To find the hole's $y$-value, substitute $x = -2$ into the simplified function: $$f(-2) = \frac{-2+1}{-2-3} = \frac{-1}{-5} = \frac{1}{5}.$$\n\n6. **Find vertical asymptotes:** The denominator of the simplified function is zero at $x = 3$, so there is a vertical asymptote at $x = 3$.\n\n7. **Find horizontal asymptotes:** For rational functions where the degrees of numerator and denominator are equal (both degree 1), the horizontal asymptote is the ratio of leading coefficients. Here, both leading coefficients are 1, so the horizontal asymptote is $$y = \frac{1}{1} = 1.$$\n\n**Final answer:**\n- Vertical asymptote at $x = 3$.\n- Hole at $x = -2$ with $y = \frac{1}{5}$.\n- Horizontal asymptote at $y = 1$.