Does a different exponent and base but same key help to resolve discrete logarithm?
Does the attacker help this knowledge about $k$ (in case (1.) or (2.)?


No; suppose you had an Oracle that could recover $a$ given $N, P, k, b, M$, then you could recover $a$ just given $N, P, k$ (that is, solve the Discrete Log problem).  Because we believe that the Discrete Log problem (given appropriate choices for $N, P$) is hard, we can conclude that constructing such an Oracle is also hard.

Here is how we would use such an Oracle; we already have been given $N, P, k$, and so all we need to do is construct $b, M$.  This can be done by selecting a $b$ relatively prime to $P-1$ and random otherwise; we can then compute $M = k^{b^{-1} \bmod P-1} \bmod P$; such a $b, M$ pair satisfies $k = M^b \mod P$, and so satisfies the Oracle's preconditions.  We then supply $N, P, k, b, M$ to the Oracle, and recover $a$

A similar method can be constructed for your question (3)