Two masses, 5 kg and 7 kg, are connected by a string over a frictionless pulley. Calculate the acceleration of the masses and the tension in the string.

Let $m_1 = 5 \, \text{kg}$ and $m_2 = 7 \, \text{kg}$.
Since $m_2 > m_1$, $m_2$ accelerates downwards and $m_1$ accelerates upwards. Let $a$ be the magnitude of the acceleration and $T$ be the tension in the string. We use $g = 9.8 \, \text{m/s}^2$.

For $m_1$ (moving upwards):

$$T - m_1 g = m_1 a \quad (1)$$

For $m_2$ (moving downwards):

$$m_2 g - T = m_2 a \quad (2)$$

$$(T - m_1 g) + (m_2 g - T) = m_1 a + m_2 a$$ $$m_2 g - m_1 g = (m_1 + m_2) a$$ $$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$

$$a = \frac{(7 \, \text{kg} - 5 \, \text{kg})(9.8 \, \text{m/s}^2)}{5 \, \text{kg} + 7 \, \text{kg}}$$ $$a = \frac{(2 \, \text{kg})(9.8 \, \text{m/s}^2)}{12 \, \text{kg}}$$ $$a = \frac{19.6 \, \text{m/s}^2}{12}$$ $$a \approx 1.63 \, \text{m/s}^2$$

The acceleration is $\boxed{\text{1.63 m/s}^2}$.

Using Equation 1:

$$T = m_1 g + m_1 a$$ $$T = m_1 (g + a)$$

$$T = 5 \, \text{kg} (9.8 \, \text{m/s}^2 + 1.6333... \, \text{m/s}^2)$$ $$T = 5 \, \text{kg} (11.4333... \, \text{m/s}^2)$$ $$T \approx 57.2 \, \text{N}$$

The tension in the string is $\boxed{\text{57.2 N}}$.

The string exerts a downward force equal to the tension $T$ on each side of the pulley. Therefore, the total downward force exerted by the string on the pulley is $2T$.
The ceiling must exert an upward force $F_c$ to support the pulley. Assuming the pulley is massless and in equilibrium (not accelerating vertically), the upward force from the ceiling must balance the total downward force from the string.

$$F_c = 2T$$

$$F_c = 2 \times 57.1666... \, \text{N}$$ $$F_c \approx 114 \, \text{N}$$

The force the ceiling must exert on the pulley is $\boxed{\text{114 N}}$.