Determine the nominal area of the triangle.

The base $b = 12$ cm and the height $h = 8$ cm.
The nominal area $A_0$ is given by the formula $A = \frac{1}{2}bh$.

$$A_0 = \frac{1}{2} \times 12 \text{ cm} \times 8 \text{ cm} = 6 \text{ cm} \times 8 \text{ cm} = 48 \text{ cm}^2$$

Since the measurements are to the nearest centimeter, the maximum possible absolute error for each measurement is half of the smallest unit, which is 1 cm.
Absolute error in base, $\Delta b = \frac{1}{2} \text{ cm} = 0.5 \text{ cm}$.
Absolute error in height, $\Delta h = \frac{1}{2} \text{ cm} = 0.5 \text{ cm}$.

a) Step 3: Calculate the absolute error in the area.
For a product of two quantities, $A = kbh$, the maximum absolute error $\Delta A$ is given by:

$$\Delta A = A_0 \left( \frac{\Delta b}{b_0} + \frac{\Delta h}{h_0} \right)$$ $$\Delta A = 48 \text{ cm}^2 \left( \frac{0.5 \text{ cm}}{12 \text{ cm}} + \frac{0.5 \text{ cm}}{8 \text{ cm}} \right)$$ $$\Delta A = 48 \left( \frac{1}{24} + \frac{1}{16} \right)$$

To add the fractions, find a common denominator, which is 48.

$$\Delta A = 48 \left( \frac{2}{48} + \frac{3}{48} \right)$$ $$\Delta A = 48 \left( \frac{2+3}{48} \right)$$ $$\Delta A = 48 \left( \frac{5}{48} \right)$$ $$\Delta A = 5 \text{ cm}^2$$

The absolute error in calculating the area of the triangle is $\boxed{5 \text{ cm}^2}$.

b) Step 4: Calculate the percentage error in the area.
The percentage error is given by the formula:

$$\text{Percentage Error} = \frac{\text{Absolute Error}}{\text{Nominal Area}} \times 100\%$$ $$\text{Percentage Error} = \frac{\Delta A}{A_0} \times 100\%$$ $$\text{Percentage Error} = \frac{5 \text{ cm}^2}{48 \text{ cm}^2} \times 100\%$$

To calculate $\frac{5}{48}$:

$$5 \div 48 \approx 0.104166...$$ $$\text{Percentage Error} \approx 0.104166... \times 100\%$$ $$\text{Percentage Error} \approx 10.4166...\%$$

Rounding to 1 decimal place:

$$\text{Percentage Error} \approx 10.4\%$$

The percentage error in the area, to 1 decimal place, is $\boxed{10.4\%}$.