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Answer 1

consider subnent mask of /28
so our network part is
192.168.16.0101_ _ _ _
now to make it directed broadcast address 
make all host bits as 1
192.168.16.0101 1 1 1 1 
which is 192.168.16.95
for limited broadcasting address make all bits as 1 so
255.255.255.255 is our limited
 

Answer 2

Identify Network and Host Bits: The address is $192.168.16.89/28$. The $/28$ indicates that the first $28$ bits are for the network and the remaining $32-28=4$ bits are for the hosts.

Find the Network Address: The network address is the first address of the subnet. To find it, we set all 4 host bits to $0$.

• The last octet is $89$, which is $01011001$ in binary.
   
• The subnet mask for $128$ in the last octet is $11110000$.
   
• Performing a bitwise AND: $01011001$ AND $11110000$ gives $01010000$.
   
• $01010000$ in decimal is $80$ . So the network address is $192.168.16.80$.

Find the Broadcast Address: To find the broadcast address, we take the network address and set all $4$ host bits to $1$.

• Network address (last octet): $01010000$.
   
• Setting the last $4$ bits to $1$ gives $01011111$.
   
• $01011111$ in decimal is $95$.
   
• Therefore, the broadcast address is $192.168.16.95$.

Comments

What does subnet mask of 128 means ? Is that a typo for /28? I'm gettting correct answer with 28.