$i)$ **DBA of the last subnet**

Starting address given to Company $= 181.55.0.0$

No. of Subnets needed $= 1000$

ie; no. of bits needed from host address part $= \left \lceil \log _2 1000 \right \rceil = 10$

The $1000$ subnets will be numbered from $0$ to $999$.

So the starting address of the $last$ subnet $= 999$,

which is $11\space 1110\space 0111$. Putting this in subnet part, the starting address of the last subnet is,

for $DBA$(Direct broadcast address), put $all\space host\space bits$ $1$.

So, $DBA$ of the $last$ subnet is,

$181.55.1111\space10 01. 1111\space 1111$

or $181.55.249.255$.

$ii)$ **Address of 4th last host of 2nd last subnet**

The starting address of the $second\space last$ subnet $(0-999)$ will be $998$, or $11\space 1110\space 0110$. Putting this in place of subnet address, we get the starting address of the $second\space last$ subnet as,

Since $All \space1$s being broadcast address, $11\space 1111$ can't be a host address, hence the $last$ host address will be $11\space 1110$.

Therefore the $last\space host$ in the above subnet is,

which is $181.55.249.190$.

So, the $fourth\space last\space host$ will have an address

$181.55.249.187$.