Halogenalkane

Shouldent it be 1-bromo-1-chloro-2,2,2-trifluoroethane

Because bromine comes first and has to be lowest

Fluorine is senior to chlorine etc. so gets the 1 position.
Bromine is first alphabetically, so comes 1st in the name.

What do you mean by senior

More important ( more reactive basically so gets priority)

Instead can I think that on carbon 1 there is 3 flourine so (3) + (2) Br +(2)Cl so 7 in total

If numbered right to left it will be (1)Cl+(1)Br +(6) F total 8

So number from left to right as lowest

F > Cl > Br > I - that is the order that their numbers are assigned (assuming there aren't any functional groups determining the numbering) - this matches your OP.
But naming is done alphabetically.

Correct me if I'm wrong, but I think it should be 2-bromo-2-chloro-1,1,1-trifluoroethane.

The bromo- comes first in the name due to alphabetical ordering, thus followed by chloro- and fluoro-.

If the carbon bonded to three fluorines is treated as carbon '1', then the total number when adding up the numbers in the name is lower, as 2+2+1+1+1=7, whereas if the fluorine-bonded carbon was carbon '2', the total number would be higher as 1+1+2+2+2=8.

I was taught that the name which gives the lowest total of numbers is the correct one to use, therefore 2-bromo-2-chloro-1,1,1-trifluoroethane would be correct.

Correct me if I'm wrong, but I think it should be 2-bromo-2-chloro-1,1,1-trifluoroethane.

The bromo- comes first in the name due to alphabetical ordering, thus followed by chloro- and fluoro-.

If the carbon bonded to three fluorines is treated as carbon '1', then the total number when adding up the numbers in the name is lower, as 2+2+1+1+1=7, whereas if the fluorine-bonded carbon was carbon '2', the total number would be higher as 1+1+2+2+2=8.

I was taught that the name which gives the lowest total of numbers is the correct one to use, therefore 2-bromo-2-chloro-1,1,1-trifluoroethane would be correct.

Correct me if I'm wrong, but I think it should be 2-bromo-2-chloro-1,1,1-trifluoroethane.

The bromo- comes first in the name due to alphabetical ordering, thus followed by chloro- and fluoro-.

If the carbon bonded to three fluorines is treated as carbon '1', then the total number when adding up the numbers in the name is lower, as 2+2+1+1+1=7, whereas if the fluorine-bonded carbon was carbon '2', the total number would be higher as 1+1+2+2+2=8.

I was taught that the name which gives the lowest total of numbers is the correct one to use, therefore 2-bromo-2-chloro-1,1,1-trifluoroethane would be correct.

It is 2-bromo-2-chloro-1,1,1-trifluoroethane, but I bet none of the UK exam boards would penalise the numbering the other way around.

CH2FCH2Cl is 2-chloro-1-fluoro-ethane. F is more senior than Cl, so takes the 1 position. Cl is earlier in the alphabet, so is written first.

CH2FCHClOH is 1-chloro-2-fluoro-ethan-1-ol. F vs Cl seniority now doesn't matter as OH is senior to both, OH gets the 1 position which determines F and Cl. Cl is earlier in the alphabet, so is written first. Since there are numbers needed, all numbers must be written hence the -1-ol bit.

Lowest total numbering is wrong - it sometimes works, but not always. The rule to use is called Point of First Difference. http://www.chem.ucalgary.ca/courses/351/WebContent/orgnom/main/difference.html

What about this example so do I label from right to left as F takes priority?

4-chloro-2-fluorobutane.

But no one would bat an eyelid at 1-chloro-3-fluorobutane (which is actually wrong).

So in exam will 1-chloro-3-flourobutane be correct