Write the equilibrium constant expression for the reaction C(s) + CO2(g) + 2Cl2(g) <=> 2COCl2(g). Calculate Kc for this reaction using the given values.

Multiply reaction (2) by 2:  
$2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g)$  
The new equilibrium constant for this reaction is $(K_{c2})^2$.

$(C(s) + CO_2(g) \rightleftharpoons 2CO(g))$  
$+$  
$(2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g))$  
Summing these gives:  
$C(s) + CO_2(g) + 2CO(g) + 2Cl_2(g) \rightleftharpoons 2CO(g) + 2COCl_2(g)$  
Canceling $2CO(g)$ from both sides yields the target reaction:  
$C(s) + CO_2(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g)$

When reactions are added, their equilibrium constants are multiplied.
$$
K_c = K_{c1} \times (K_{c2})^2
$$
$$
K_c = (1.4 \times 10^{12}) \times (5.5 \times 10^1)^2
$$
$$
K_c = (1.4 \times 10^{12}) \times (30.25 \times 10^2)
$$
$$
K_c = (1.4 \times 10^{12}) \times (3.025 \times 10^3)
$$
$$
K_c = (1.4 \times 3.025) \times 10^{12+3}
$$
$$
K_c = 4.235 \times 10^{15}
$$
The equilibrium constant for the reaction is $\boxed{\mathbf{4.235 \times 10^{15}}}$.

This reaction is the reverse of the original reaction.

When a reaction is reversed, the new equilibrium constant is the reciprocal of the original.
$$
K_p' = \frac{1}{K_p}
$$
$$
K_p' = \frac{1}{7.1 \times 10^4}
$$
$$
K_p' \approx 1.408 \times 10^{-5}
$$
The value of $K_p$ is $\boxed{\mathbf{1.408 \times 10^{-5}}}$.

This reaction is the original reaction multiplied by a factor of $\frac{1}{2}$.

When a reaction is multiplied by a factor 'n', the new equilibrium constant is the original constant raised to the power 'n'. Here, $n = \frac{1}{2}$.
$$
K_p'' = (K_p)^{1/2} = \sqrt{K_p}
$$
$$
K_p'' = \sqrt{7.1 \times 10^4}
$$
$$
K_p'' = \sqrt{71000}
$$
$$
K_p'' \approx 266.458
$$
The value of $K_p$ is $\boxed{\mathbf{266.458}}$.