Subgroups of finitary symmetric groups

The answer to Question 2 is "yes". Take the relatively free group with law $x^4=1$ and infinite set of generators. That group is locally finite (Sanov, I. N. Solution of Burnside's problem for exponent 4. Leningrad State Univ. Annals [Uchenye Zapiski] Math. Ser. 10, (1940). 166–170.). It is not inside the group of finitary permutations (every permutation of order 4 is a product of 4-cycles and 2-cycles) and no non-identity element has roots of order 4.

Update It is even easier to consider the relatively free group $G$ of exponent $3$ instead. That group is solvable (M. Hall, The Theory of Groups, pages 320-324), every non-zero element does not have a root of degree 3, and is not inside $S_\infty$, the group of finitary permutations of an infinite set. Indeed, suppose $G$ is inside $S_\infty$. We can assume that it is transitive (exercise). But this contradicts Corollary 4.7 in the notes cited by Igor Rivin (Wiegold's theorem).

Well, for the first question, there is the amazing result that the only simple infinite such group is the finitary alternating group, see Chris Pinnock's notes (the Mihles-Tyskevic theorem). That tells you that the theory is not too much richer than the theory of finite groups.

The notes at my site have moved to here: Finitary Permutation Groups