LL(1) grammar and first & follow sets

Question

I think the below is actually LL(1) but I am not 100% sure. are we able to prove that this iss LL(1) grammar and then also, is the first and follow sets given correct? I dont really understand for sure how to actually get the follow sets.

Grammar

expr::= term {addop term}
term::= factor {mulop factor}
factor::= variable| unsigned-number| ( expr)
variable::= identifier
addop::= + | -
mulop::= * | /

First Sets

First(expr) = identifier, unsigned number, (
First(Term) = identifier, unsigned number, (
First(Factor) = identifier, unsigned number, (
First(Variable) = Identifier
First(addop) = +, -
First(mulop) = *, /

Follow Sets

Follow(Expr) = First(term)
Follow(Term) = First(Factor)

Do not understand how to do the Follow sets of the below sets
Follow(Factor) = First( ")" ) = )
Follow(Variable) Follow(addop) Follow(mulop)

Answer 1

FIRST sets are correct.

FOLLOW sets:

FOLLOW(A) of non-terminal A is the set of terminal symbols that can follow in the derivation sequence

FOLLOW(expr): check where it appeared in the right-hand side of production. It is there is factor:=(expr), when we take this production in the derivation what follows expr is ) and expr is a start symbol.

   FOLLOW(expr)={),$}

similarly,

   FOLLOW(term) = FIRST(addop) U FOLLOW(expr) = {+,-,),$}
   FOLLOW(factor) = FIRST(mulop) U FOLLOW(trem) = {*,/,+,-,),$}
   FOLLOW(addop) = FIRST(term) = {identifier,unsigned number,(}
   FOLLOW(multop) = FIRST(factor) = {identifier,unsigned number,(}
   FOLLOW(Variable)= FOLLOW(factor) = {*,/,+,-,),$}

where $ is the end of the input.

Given grammar is LL(1). FIRST sets of factor are disjoint and we don't need check the FOLLOW sets as there are no epsilon productions.