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, 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).

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).

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.

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, 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).