In a triangle ABC, AB=3, BC=4, and angle ABC=90 degrees. AD=13 and angle ACD=90 degrees. Find tan(angle DAC).

Given $AB = 3$ cm and $BC = 4$ cm, and $\angle ABC = 90^\circ$.

$$AC^2 = AB^2 + BC^2$$ $$AC^2 = (3\, \text{cm})^2 + (4\, \text{cm})^2$$ $$AC^2 = 9\, \text{cm}^2 + 16\, \text{cm}^2$$ $$AC^2 = 25\, \text{cm}^2$$ $$AC = \sqrt{25\, \text{cm}^2}$$ $$AC = 5\, \text{cm}$$

Given $AD = 13$ cm and $AC = 5$ cm (from Step 1), and $\angle ACD = 90^\circ$.

$$AD^2 = AC^2 + CD^2$$ $$(13\, \text{cm})^2 = (5\, \text{cm})^2 + CD^2$$ $$169\, \text{cm}^2 = 25\, \text{cm}^2 + CD^2$$ $$CD^2 = 169\, \text{cm}^2 - 25\, \text{cm}^2$$ $$CD^2 = 144\, \text{cm}^2$$ $$CD = \sqrt{144\, \text{cm}^2}$$ $$CD = 12\, \text{cm}$$

In the right-angled triangle $\triangle ACD$, for angle $DAC$:
The opposite side is $CD = 12$ cm.
The adjacent side is $AC = 5$ cm.
The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side.

$$\tan(\angle DAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CD}{AC}$$ $$\tan(\angle DAC) = \frac{12\, \text{cm}}{5\, \text{cm}}$$ $$\tan(\angle DAC) = \frac{12}{5}$$

The value of $\tan(\angle DAC)$ is $\boxed{\frac{12}{5}}$.