Irska

a. For part (i), first observe that if $A(i,j)=M$ with $i,j>1$, then the three neighbouring values that have average value $A(i,j)$ must also equal $M$, since otherwise at least one would have to be larger than $M$. Then it is not hard to see that we can propagate this value to get that the subarray with main diagonal from position $(1,1)$ to $(i,j)$ must equal $M$ everywhere, thus yielding (i).

b. For (ii), it already follows from the previous paragraph that there are no non-trivial examples if $i_0=j_0=n$. However, such non-trivial examples can be constructed in all other cases: just pick $A(i,j)=M$ for all numbers in the subarray with main diagonal from $(1,1)$ to $(i_0,j_0)$. This leaves some elements $A(i,1)$ undefined (if $i_0<n$) or some elements $A(1,i)$ undefined (if $j_0<n$). Put numbers strictly less than $M$ in all positions in the first row and first column that are not yet defined. Now fill in the rest of the array one row at a time from left to right, beginning with the first row and proceeding in the natural order. In all cases, the next position is uniquely defined in terms of the values we have already chosen. This is the required example.