A) IPv4 and IPv6 are both Internet Protocol versions used to identify devices on a network.

We need $2^h - 2 \ge 52 \implies 2^h \ge 54$.
The smallest power of 2 greater than or equal to 54 is $2^6 = 64$.
So, $h=6$ host bits.
The subnet mask is $32 - 6 = /26$.
In dotted decimal: $255.255.255.(256 - 2^6) = 255.255.255.(256 - 64) = 255.255.255.192$.
The subnet mask for Department A is $\boxed{255.255.255.192}$.

We need $2^h - 2 \ge 29 \implies 2^h \ge 31$.
The smallest power of 2 greater than or equal to 31 is $2^5 = 32$.
So, $h=5$ host bits.
The subnet mask is $32 - 5 = /27$.
In dotted decimal: $255.255.255.(256 - 2^5) = 255.255.255.(256 - 32) = 255.255.255.224$.
The subnet mask for Department B is $\boxed{255.255.255.224}$.

We need $2^h - 2 \ge 28 \implies 2^h \ge 30$.
The smallest power of 2 greater than or equal to 30 is $2^5 = 32$.
So, $h=5$ host bits.
The subnet mask is $32 - 5 = /27$.
In dotted decimal: $255.255.255.(256 - 2^5) = 255.255.255.(256 - 32) = 255.255.255.224$.
The subnet mask for Department C is $\boxed{255.255.255.224}$.

We need $2^h - 2 \ge 14 \implies 2^h \ge 16$.
The smallest power of 2 greater than or equal to 16 is $2^4 = 16$.
So, $h=4$ host bits.
The subnet mask is $32 - 4 = /28$.
In dotted decimal: $255.255.255.(256 - 2^4) = 255.255.255.(256 - 16) = 255.255.255.240$.
The subnet mask for Department D is $\boxed{255.255.255.240}$.