As for the first question, a backtracking algorithm https://www.staff.science.uu.nl/~kalle101/sixy.c shows there are 1936 completions of
$$\matrix{1&2&3&4&5&6\cr *&*&*&*&*& *\cr *&*&*&*&*&*\cr *&*&*&*&*&*\cr *&*&*&*&*&*\cr *&*&*&*&*&*&}$$
The third question does not seem to have a pleasant answer. For instance, there are 2752 ways to chose five cells in
$$\matrix{ *&*&*&*&*& *\cr *&*&*&*&*& *\cr *&*&6&*&*&*\cr *&*&*&*&*&*\cr *&*&*&*&6&*\cr *&*&*&*&*&*}$$ to construct an initial clue like $$\matrix{ *&*&*&*&*& *\cr *&*&*&*&*& 1\cr *&*&6&*&2&*\cr *&3&*&*&*&*\cr *&4&*&*&6&5\cr *&*&*&*&*&*}$$ that ensures a unique solution. Here the construction fills the chosen cells with 1 through 5 consistent with the lexicographic order. https://www.staff.science.uu.nl/~kalle101/sixy5.c
As for the second question, the answer is $n^*=7$. One sees this by replacing $$\matrix{ *&*&*&*&*& *\cr *&*&*&*&*& *\cr *&*&6&*&*&*\cr *&*&*&*&*&*\cr *&*&*&*&6&*\cr *&*&*&*&*&*}$$ with starting configurations that have just one filled cell. There is a symmetry group of order 128 with just three orbits of cells, and we tried all 6 times 3 orbits of configurations that have just one filled cell.
To answer the third question in full, one could try all orbits of configurations that have two filled cells. And then there will be double counts. For instance $$\matrix{ 0& 0& 0& 0& 0& 0 \cr 0& 1 & 0& 0& 2& 0 \cr 0& 2& 0& 0& 0& 3 \cr 0& 0& 0& 0& 0& 0 \cr 0& 0& 0& 0& 0& 0 \cr 4& 0& 0& 0& 5& 2 }$$ would be found starting from $$\matrix{ 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 0\cr 0& 2& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 2}$$ and starting from $$\matrix{ 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 2& 0\cr 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 0\cr 0& 0& 0& 0& 0& 2}$$ It is not clear that such investigation is worthwhile.