If tan A = 1/2 and cos B = 3/5, where both A and B are acute angles, what is the exact value of sin(A + B)?

Since $A$ is an acute angle, we can form a right-angled triangle where the opposite side is 1 and the adjacent side is 2.
The hypotenuse $h_A$ is calculated using the Pythagorean theorem:

$$h_A = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$

Therefore,

$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}$$ $$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}$$

Since $B$ is an acute angle, we can form a right-angled triangle where the adjacent side is 3 and the hypotenuse is 5.
The opposite side $o_B$ is calculated using the Pythagorean theorem:

$$o_B = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$$

Therefore,

$$\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}$$

Substitute the values found in Step 1 and Step 2:

$$\sin(A+B) = \left(\frac{1}{\sqrt{5}}\right) \left(\frac{3}{5}\right) + \left(\frac{2}{\sqrt{5}}\right) \left(\frac{4}{5}\right)$$ $$\sin(A+B) = \frac{3}{5\sqrt{5}} + \frac{8}{5\sqrt{5}}$$ $$\sin(A+B) = \frac{3+8}{5\sqrt{5}}$$ $$\sin(A+B) = \frac{11}{5\sqrt{5}}$$

This matches option (a).

The final answer is $\text{(a)} \frac{11}{5\sqrt{5}}$.