Netmask in the video is 255.255.178.0 and gateway 192.168.178.1?
Using CIDR notation 255.255.255.0 is /24, meaning 24 bits are locked. You can do your subnetting any way fits you. If you need more than 254 addresses in your LAN, you could use a /16 bit netmask, and have 65534 addresses available. But the mask would be 255.255.0.0
You could use a /17 bit netmask and that would leave half the addresses available on a /16 subnet, but netmask would be 255.255.128.0
A /18 bit netmask would be 255.255.192.0
I'm not aware that netmask could use fractions of a bit. So that netmask is wrong. Perhaps they have a typo and it would be 255.255.128.0 which corresponds to a 17 bit netmask. If you need 32766 IP addresses available in your LAN, that's the way to do it.
Let's see what ipcalc says. We are going to use gateway' address and netmask of 17 bits to see what the resultant networks are:
tatel
@gis:~$ ipcalc 192.168.178.1/17
Address: 192.168.178.1 11000000.10101000.1 0110010.00000001
Netmask: 255.255.128.0 = 17 11111111.11111111.1 0000000.00000000
Wildcard: 0.0.127.255 00000000.00000000.0 1111111.11111111
=>
Network: 192.168.128.0/17 11000000.10101000.1 0000000.00000000
HostMin: 192.168.128.1 11000000.10101000.1 0000000.00000001
HostMax: 192.168.255.254 11000000.10101000.1 1111111.11111110
Broadcast: 192.168.255.255 11000000.10101000.1 1111111.11111111
Hosts/Net: 32766 Class C, Private Internet
So, even if that netmask would be the right one 255.255.128.0, which it isn't, this device will be able to speak only to other devices with IP address between 192.168.128.1 and 192.168.191.254. Quite probably you LAN is out of that range?
I think, you need to change that device network configuration and adapt it to your LAN configuration, as explained in the previous post. It should work with the usual 255.255.255.0 netmask as long as device's IP is in the address range of the LAN configured in your router. If not, get into your router via web browser, take some screenshots of your LAN configuration, and send it to these guys.