To find the earliest time all 120 parts can be finished, we must determine the optimal allocation of parts between Machine A and Machine B. Since both machines operate simultaneously, the total time to finish is determined by the machine that takes the longest to complete its assigned share of the work.

Let $N_A$ be the number of parts produced by Machine A.
Let $N_B$ be the number of parts produced by Machine B.

The constraints are:
1.  $N_A + N_B = 120$
2.  $N_A, N_B$ must be non-negative integers.
3.  $N_B = 120 - N_A$

### Step 1: Define Time Functions

**Machine A:**
*   Setup time: 30 minutes = $0.5$ hours.
*   Production rate: 8 parts/hour.
*   Time to produce $N_A$ parts: $T_A(N_A) = 0.5 + \frac{N_A}{8}$ hours.

**Machine B:**
*   Setup time: 12 minutes = $\frac{12}{60}$ hours = $0.2$ hours.
*   Production rate: 5 parts/hour.
*   Time to produce $N_B$ parts: $T_B(N_B) = 0.2 + \frac{N_B}{5}$ hours.

The total time to finish all parts is $T_{total} = \max(T_A(N_A), T_B(120 - N_A))$. We want to minimize this maximum time.

### Step 2: Find the Continuous Optimal Allocation

First, let's find the allocation where the completion times are equal, ignoring the integer constraint for a moment. This gives us the theoretical lower bound.

Set $T_A(N_A) = T_B(120 - N_A)$:

$$0.5 + \frac{N_A}{8} = 0.2 + \frac{120 - N_A}{5}$$

Subtract 0.2 from both sides:
$$0.3 + \frac{N_A}{8} = \frac{120 - N_A}{5}$$

Multiply the entire equation by 40 (the least common multiple of 8 and 5) to clear denominators:
$$40(0.3) + 40\left(\frac{N_A}{8}\right) = 40\left(\frac{120 - N_A}{5}\right)$$
$$12 + 5N_A = 8(120 - N_A)$$
$$12 + 5N_A = 960 - 8N_A$$

Add $8N_A$ to both sides and subtract 12 from both sides:
$$13N_A = 948$$
$$N_A = \frac{948}{13} \approx 72.92$$

Since $N_A$ must be an integer, we must check the integers closest to 72.92, which are **72** and **73**.

### Step 3: Check Integer Completions

We calculate the total time for $N_A = 72$ and $N_A = 73$.

**Case 1: $N_A = 72$**
*   $N_B = 120 - 72 = 48$
*   Time for Machine A:
    $$T_A = 0.5 + \frac{72}{8} = 0.5 + 9 = 9.5 \text{ hours}$$
*   Time for Machine B:
    $$T_B = 0.2 + \frac{48}{5} = 0.2 + 9.6 = 9.8 \text{ hours}$$
*   Total Time: $\max(9.5, 9.8) = \mathbf{9.8}$ **hours**.

**Case 2: $N_A = 73$**
*   $N_B = 120 - 73 = 47$
*   Time for Machine A:
    $$T_A = 0.5 + \frac{73}{8} = 0.5 + 9.125 = 9.625 \text{ hours}$$
*   Time for Machine B:
    $$T_B = 0.2 + \frac{47}{5} = 0.2 + 9.4 = 9.6 \text{ hours}$$
*   Total Time: $\max(9.625, 9.6) = \mathbf{9.625}$ **hours**.

**Case 3: Check $N_A = 74$ (to ensure we didn't miss a better integer further away)**
*   $N_B = 120 - 74 = 46$
*   Time for Machine A:
    $$T_A = 0.5 + \frac{74}{8} = 0.5 + 9.25 = 9.75 \text{ hours}$$
*   Time for Machine B:
    $$T_B = 0.2 + \frac{46}{5} = 0.2 + 9.2 = 9.4 \text{ hours}$$
*   Total Time: $\max(9.75, 9.4) = \mathbf{9.75}$ **hours**.

**Case 4: Check $N_A = 71$ (to ensure we didn't miss a better integer further away)**
*   $N_B = 120 - 71 = 49$
*   Time for Machine A:
    $$T_A = 0.5 + \frac{71}{8} = 0.5 + 8.875 = 9.375 \text{ hours}$$
*   Time for Machine B:
    $$T_B = 0.2 + \frac{49}{5} = 0.2 + 9.8 = 10.0 \text{ hours}$$
*   Total Time: $\max(9.375, 10.0) = \mathbf{10.0}$ **hours**.

### Step 4: Analysis and Conclusion

Comparing the results:
*   $N_A = 71 \rightarrow 10.0$ hours
*   $N_A = 72 \rightarrow 9.8$ hours
*   $N_A = 73 \rightarrow 9.625$ hours
*   $N_A = 74 \rightarrow 9.75$ hours

The minimum time occurs at $N_A = 73$.

**Verification of "Immediately Preceding Completion Time":**
The prompt asks to check the immediately preceding completion time. In the context of discrete integer allocations, this usually refers to checking the neighbor integers ($N_A=72$ and $N_A=74$) to confirm the local minimum. We have done this above.
*   At $N_A=72$, Machine B is the bottleneck (9.8h).
*   At $N_A=73$, Machine A is the bottleneck (9.625h).
*   At $N_A=74$, Machine A is the bottleneck (9.75h).

The transition from $N_A=72$ to $N_A=73$ shifts the bottleneck from Machine B to Machine A. The time decreases from 9.8h to 9.625h. The transition from $N_A=73$ to $N_A=74$ keeps the bottleneck on Machine A, but increases the time because Machine A is slower per part than Machine B relative to the shift in load? Let's check the slopes.
Slope of $T_A$ w.r.t $N_A$ is $1/8 = 0.125$.
Slope of $T_B$ w.r.t $N_A$ (since $N_B$ decreases as $N_A$ increases) is $-1/5 = -0.2$.
At $N_A=73$, $T_A > T_B$. Increasing $N_A$ further increases $T_A$ by 0.125 per unit and decreases $T_B$ by 0.2 per unit. Since $T_A$ is the max, increasing $N_A$ increases the max time. Decreasing $N_A$ to 72 decreases $T_A$ by 0.125 but increases $T_B$ by 0.2. Since $T_B$ was lower than $T_A$ at 73? No, at 73, $T_A=9.625$ and $T_B=9.6$. $T_A$ is the bottleneck.
If we go to 72: $T_A$ drops to 9.5, $T_B$ rises to 9.8. Max is 9.8.
If we go to 74: $T_A$ rises to 9.75, $T_B$ drops to 9.4. Max is 9.75.

Thus, 9.625 is indeed the minimum.

**Final Answer Calculation:**
The earliest time is **9.625 hours**.

To express this in hours and minutes:
$0.625 \text{ hours} = 0.625 \times 60 \text{ minutes} = 37.5 \text{ minutes}$.

**Answer:**
The earliest time all 120 parts can be finished is **9 hours and 37.5 minutes** (or **9.625 hours**).