Solution

• L3 = L1 ∩ L2
  TRUE. L1 and L2 will have only those strings in common which will have equal number of 0s then equal number of 1s and then equal number of 0s. This is exactly what L3 is saying, equal number of 0s then 1s and then 0s.
   
• L1 and L2 are context free languages but L3 is not a context free language.
  TRUE. It is saying in L1 that the number of 0s present before 1 should be equal to the total number of 1s,which is possible with one stack. L2 is saying that the number of 1s present before 0 should be equal to the total number of 0s,which is possible with one stack. And L3 is saying that the number of 0s and then 1s and then 0s should be equal, which is not possible using one stack.
   
• L1 and L2 are not context free languages but L3 is a context free language.
  FALSE. L1 and L2 are context free languages and L2 is not context free language. It is saying in L1 that the number of 0s present before 1 should be equal to the total number of 1s. L2 is saying that the number of 1s present before 0 should be equal to the total number of 0s. And L3 is saying that the number of 0s and then 1s and then 0s should be equal, which is not possible using one stack.
  
• L1 is a subset of L3.
  FALSE. Reverse is true. L3 is the subset of L1. L1 is a larger set having all the elements of L3, not the vice versa.