The answer is $\sum_{j=0}^{k+1}{n\choose j}$ and is realised by all subsets containing at most $k+1$ elements of $M$. Two such distinct subsets have indeed at most $k$ elements in common.

Let now $\mathcal S$ be a set of subsets in $M$ satisfying your requirement.

Suppose $A\in S$ is a subset of cardinality $>k+1$ intersecting a set $B\in S\setminus A$ in $j\leq k$ elements. Suppose $j$ maximal, ie $\sharp (A\cap C)\leq j$ for all $C\in S\setminus A$. Replace $A$ by $\tilde A=A\cap B\cup \lbrace a_1,\dots,a_{k+1-j}\rbrace\subset A$ with $a_1, \dots,a_{k+1-j}\in A\setminus B$. The map $A\longmapsto \tilde A$ replaces $A$ by a subset containing $k+1$ elements which does not belong to $S$. Moreover, $\tilde S=(S\cup \tilde A)\setminus A$ satisfies your requirements. Iterating this construction leads to a set $S$ of the same cardinality consisting only of subsets with at most $k+1$ elements.

The answer (to question 1) is $\sum_{j=0}^{k+1}{n\choose j}$ and is realised by all subsets containing at most $k+1$ elements of $M$. Two such distinct subsets have indeed at most $k$ elements in common.

Let now $\mathcal S$ be a set of subsets in $M$ satisfying your requirement.

Suppose $A\in S$ is a subset of cardinality $>k+1$ intersecting a set $B\in S\setminus A$ in $j\leq k$ elements. Suppose $j$ maximal, ie $\sharp (A\cap C)\leq j$ for all $C\in S\setminus A$. Replace $A$ by $\tilde A=A\cap B\cup \lbrace a_1,\dots,a_{k+1-j}\rbrace\subset A$ with $a_1, \dots,a_{k+1-j}\in A\setminus B$. The map $A\longmapsto \tilde A$ replaces $A$ by a subset containing $k+1$ elements which does not belong to $S$. Moreover, $\tilde S=(S\cup \tilde A)\setminus A$ satisfies your requirements. Iterating this construction leads to a set $S$ of the same cardinality consisting only of subsets with at most $k+1$ elements.

Concerning question 2, the above proof gives the inequality $N(n,k,m)\leq {n\choose k+1}$ if $m>k+1$ (and $N(n,k,m)={n\choose m}$ if $m\leq k+1$).

The answer is $\sum_{j=0}^{k+1}{n\choose j}$ and is realised by all subsets containing at most $k+1$ elements of $M$. Two such distinct subsets have indeed at most $k$ elements in common.

Let now $\mathcal S$ be a set of subsets in $M$ satisfying your requirement.

Suppose $A\in S$ is a subset of cardinality $>k+1$ intersecting a set $B\in S\setminus A$ in $j\leq k$ elements. Suppose $j$ maximal, ie $\sharp (A\cap C)\leq j$ for all $C\in S\setminus A$. Replace $A$ by $\tilde A=A\cap B\cup \lbrace a_1,a_{k+1-j}\rbrace\subset A$ with $a_1,\dots,a_{k+1-j}\in A\setminus B$. The map $A\longmapsto \tilde A$ replaces $A$ by a subset containing $k+1$ elements which does not belong to $S$. Moreover, $\tilde S=(S\cup \tilde A)\setminus A$ satisfies your requirements. Iterating this construction leads to a set $S$ of the same cardinality consisting only of subsets with at most $k+1$ elements.

The answer is $\sum_{j=0}^{k+1}{n\choose j}$ and is realised by all subsets containing at most $k+1$ elements of $M$. Two such distinct subsets have indeed at most $k$ elements in common.

Let now $\mathcal S$ be a set of subsets in $M$ satisfying your requirement.

Suppose $A\in S$ is a subset of cardinality $>k+1$ intersecting a set $B\in S\setminus A$ in $j\leq k$ elements. Suppose $j$ maximal, ie $\sharp (A\cap C)\leq j$ for all $C\in S\setminus A$. Replace $A$ by $\tilde A=A\cap B\cup \lbrace a_1,\dots,a_{k+1-j}\rbrace\subset A$ with $a_1, \dots,a_{k+1-j}\in A\setminus B$. The map $A\longmapsto \tilde A$ replaces $A$ by a subset containing $k+1$ elements which does not belong to $S$. Moreover, $\tilde S=(S\cup \tilde A)\setminus A$ satisfies your requirements. Iterating this construction leads to a set $S$ of the same cardinality consisting only of subsets with at most $k+1$ elements.