Let $G$ be a discrete group and K a subgroup of G . denote by $(\hat{H_K})^i$ the bounded cohomology groups of $K$ , and by $(\hat{H_G})^i$ the bounded cohomology groups of G.
then $(\hat{H_K})^i$ is embedded in $(\hat{H_G})^i$ ??
Let $G$ be a discrete group and K a subgroup of G . denote by $(\hat{H_K})^i$ the bounded cohomology groups of $K$ , and by $(\hat{H_G})^i$ the bounded cohomology groups of G. then $(\hat{H_K})^i$ is embedded in $(\hat{H_G})^i$ ?? 


In general, no. There is not even a natural map (in general) in the direction you want. There is a natural map in the opposite direction, namely restriction, and this is sometimes an embedding, but not always. See chapter 8.6 in "Continuous bounded cohomology of locally compact groups", Lecture Notes in Mathematics 1758 

