Do all mixed strategy Nash Equilibria satisfy Trembling Hand perfection? Moreover, does it fulfill other criteria automatically, like Myerson's "Proper equilibrium" refinement?
No. Trembling hand perfection would be an additional consideration.

Consider this case:

Two agents engage in a two stage game. In stage one, player one selects Rock, Paper or Scissors. In stage two, player two selects Rock, Paper or Scissors. They receive payouts depending on whether or not they match. They both have identical payout matrices.

Payoffs (Match, Unmatched)


Rock (1,0)
Paper (1,0) 
Scissors (1,-1)


In a mixed strategy nash equillibrium consider player one. He can choose any mixed strategy such that

p(Rock)+p(Paper)+p(Scissors)=1

and they would all be equally rational. Since even if he chooses scissors, a rational player two would match him. However, taking into account trembling hand perfection, the subset of strategies where

p(Scissors)=0

dominate the previous strategy set. This is the case because we are concerned that player two has a chance of making a mistake if we choose scissors and not matching us. Therefore, mixed strategy and trembling hand mixed strategy give us different sets of strategies which fulfill our Nash equilibrium criteria.