Yes. If you have a linear operator, L, that acts on a vector space V, L:V\rightarrow V, we can define a representation in the following manner. Choose an orthonormal basis \{ \mathbf{v}_a \} of V. For convenience, we'll use the index a to label the basis elements, so we're implicitly treating this as if the vector space is finite. Now the action of L on the basis vectors can be decomposed in terms of the basis vectors as
L( \mathbf{v}_a ) =\sum_b L_{ba} \mathbf{v}_b.
The matrix L_{ab} is called the representation of L on the vector space V. I believe that I've chosen the order of indices correctly so that it has the proper action on a vector w = \sum_c \mathbf{v}_c when w is expressed as a column vector.
When we have a collection of operators L_i, you can show that the matrices that furnish their representation satisfy the same algebra. For your angular momentum case, you can understand the Pauli matrices as a representation.
Not quite. The operators J_i already satisfy a Lie algebra. Exponentiating them would give us the Lie group, but we don't have to do that.
So can you use any physical or other intuition to say something about states that have different j quantum numbers in the vector space V? You will probably want to recall everything that you know about the angular momentum states, like eigenvalues/vectors and ladder operators.