1. **State the problem:** We are given two functions $f(x) = x^2 - 8x + 7$ and $g(x) = x - 1$. We need to find the values of $(f + g)(0)$, $(f - g)(1)$, $(f \cdot g)(2)$, and $\left(\frac{f}{g}\right)(3)$.\n\n2. **Recall the definitions:**\n- $(f + g)(x) = f(x) + g(x)$\n- $(f - g)(x) = f(x) - g(x)$\n- $(f \cdot g)(x) = f(x) \times g(x)$\n- $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided $g(x) \neq 0$.\n\n3. **Calculate each value:**\n\n- $(f + g)(0) = f(0) + g(0)$\nCalculate $f(0)$: $$f(0) = 0^2 - 8 \times 0 + 7 = 7$$\nCalculate $g(0)$: $$g(0) = 0 - 1 = -1$$\nSo, $$(f + g)(0) = 7 + (-1) = 6$$\n\n- $(f - g)(1) = f(1) - g(1)$\nCalculate $f(1)$: $$f(1) = 1^2 - 8 \times 1 + 7 = 1 - 8 + 7 = 0$$\nCalculate $g(1)$: $$g(1) = 1 - 1 = 0$$\nSo, $$(f - g)(1) = 0 - 0 = 0$$\n\n- $(f \cdot g)(2) = f(2) \times g(2)$\nCalculate $f(2)$: $$f(2) = 2^2 - 8 \times 2 + 7 = 4 - 16 + 7 = -5$$\nCalculate $g(2)$: $$g(2) = 2 - 1 = 1$$\nSo, $$(f \cdot g)(2) = -5 \times 1 = -5$$\n\n- $\left(\frac{f}{g}\right)(3) = \frac{f(3)}{g(3)}$\nCalculate $f(3)$: $$f(3) = 3^2 - 8 \times 3 + 7 = 9 - 24 + 7 = -8$$\nCalculate $g(3)$: $$g(3) = 3 - 1 = 2$$\nSo, $$\left(\frac{f}{g}\right)(3) = \frac{-8}{2} = -4$$\n\n4. **Final answers:**\n- $(f + g)(0) = 6$\n- $(f - g)(1) = 0$\n- $(f \cdot g)(2) = -5$\n- $\left(\frac{f}{g}\right)(3) = -4$