A U-shaped tube contains water and mercury. The water column is 15 cm high. Calculate the gauge pressure at the interface and the vertical distance h to the mercury's top.

The gauge pressure at a depth $h$ in a fluid is given by $P = \rho g h$.

$$P_{gauge} = \rho_w g h_w$$ $$P_{gauge} = (1000 \text{ kg/m}^3)(10 \text{ m/s}^2)(0.15 \text{ m})$$ $$P_{gauge} = 1500 \text{ Pa}$$

The gauge pressure at the water-mercury interface is $\boxed{\text{1500 Pa}}$.

At the same horizontal level (the water-mercury interface in the left arm), the pressure must be equal. Let $2x$ be the height of the mercury column in the right arm above this interface.
The pressure exerted by the water column in the left arm is equal to the pressure exerted by the mercury column of height $2x$ in the right arm.

$$\rho_w g h_w = \rho_{Hg} g (2x)$$ $$h_w \rho_w = 2x \rho_{Hg}$$

$$2x = \frac{h_w \rho_w}{\rho_{Hg}}$$ $$2x = \frac{(0.15 \text{ m})(1000 \text{ kg/m}^3)}{13600 \text{ kg/m}^3}$$ $$2x = \frac{150}{13600} \text{ m}$$ $$2x \approx 0.011029 \text{ m}$$

The distance $h$ is the difference between the height of the water column ($h_w$) and the height of the mercury column ($2x$) above the water-mercury interface.

$$h = h_w - 2x$$ $$h = 0.15 \text{ m} - 0.011029 \text{ m}$$ $$h = 0.138971 \text{ m}$$

Convert to centimeters and round to one decimal place:

$$h \approx 13.9 \text{ cm}$$

The vertical distance $h$ is $\boxed{\text{13.9 cm}}$.