$(1)$ According to the given data, the maximum range of addresses possible in each interfaces are:
Interface $A\Rightarrow 255.255.255.128$ to $255.255.255.191$
Interface $B\Rightarrow 255.255.255.192$ to $255.255.255.223$
Interface $C\Rightarrow 255.255.255.224$ to $255.255.255.255$
Clearly, the pair of IP addresses which belong to Interface $B$ is $200.200.200.200$ & $200.200.200.205$,
which is option $(B)$
$(2)$
Interface $A \Rightarrow 255.255.255.128 = 255.255.255.100\space 0\space 0000$
Interface $B \Rightarrow 255.255.255.192 = 255.255.255.110\space 0\space 0000$
Interface $C \Rightarrow 255.255.255.224 = 255.255.255.111 \space 0\space 0000$
Looking at the subnet masks, we can infer that first 3 of the host bits are used for subnetting. So maximum possible number of subnets in this configuration is $2^3 = 8.$
So, the answer is $(C)$