First of all, a heuristic is said to be admissible if and only if $h(n)\leq h^*(n)$ for every state $n$, where $h(n)$ is your heuristic function and $h^*(n)$ is the cost of an optimal path from $n$ to the goal.
Now, when considering the sliding-tile puzzle, the Manhattan distance can be shown to be an admissible heuristic function as it results from a constraint relaxation .
From here, consider your first question: if the Manhattan distance is an admissible heuristic function for the sliding-tile puzzle where the cost of each movement is 1, it is also an admissible heuristic function for the case where the cost of each movement is strictly greater or equal than 1 (because all tiles are numbered 1 onwards).
Regarding your second question, a heuristic function $h_1(n)$ is said to dominate another heuristic function $h_2(n)$ if and only if $h_1(n)\geq h_2(n)$ for all states $n$. For some restrictive models (a unique goal, constant branching factor, no duplicates, etc.) it can be shown that more informed heuristics (i.e., $h_1(n)$ in the previous definition) expand less nodes when using A$^*$. From here, if you enhance your heuristic function with the average of all tiles, then it is necessarily more informed than pure Manhattan but watch out, it should be easy for you to note that this second heuristic function is not necessarily admissible as it can very easily overestimate the effort to reach the goal. Thus, using this second heuristic it is very likely that you will expand less nodes but you cannot guarantee that the solution returned is optimal.
Hope this helps,
 Pearl, Judea. Heuristics - Intelligent Search Strategies for Computer Problem Solving. Addison Wesley, 1984.