The answer to the first question is clearly _no_ , since the mapping can collapse the domain to the zero vector. A function that simply interchanges two points has an inverse that takes bounded sets to bounded sets, but the function is neither continuous nor linear.
The second statement is precisely equivalent to the first, so it does indeed mean that the image of a bounded set under a continuous linear mapping is bounded. The answer to the final question is _no_ , just as in the first part.