**1.** "Non-invertible" means rank $\leq n-1$, and thus the upper bound $n\left(n-1\right)$ follows from the Theorem in paragraph 8.3 in Victor Prasolov's "Problems and theorems in linear algebra". (Scroll to page 58.) The reference given there is
Flanders H., *On spaces of linear transformations with bound rank*, J. London Math. Soc. 37 (1962), pp. 10-16.

**2.** We can WLOG assume that our subspace $N$ is actually a subalgebra of $\mathrm{M}_n\left(\mathbb C\right)$ (because otherwise, we can replace it by the subalgebra it generates, and it will still have the property that any two of its elements commute), so the question is how large a commutative subalgebra of $\mathrm{M}_n\left(\mathbb C\right)$ can get. The answer is that the maximum possible dimension of such a subalgebra is $\left\lfloor \dfrac{n^2}{4} \right\rfloor + 1$, and this is a result of I. Schur (see the 2 links in that topic). A (relatively) short proof can be found in M. Mirzakhani, *A Simple Proof of a Theorem of Schur*, The American Mathematical Monthly, Vol. 105, No. 3 (Mar., 1998), pp. 260--262.

**4.** Here the maximal dimension is $1$, and Petya has told why.

As for **3.**, I can prove the upper bound $\frac{n^2}{2}$ (strangely enough, for $\mathbb C$ only), but unfortunately there is room between it and the lower bound $\frac{n\left(n-1\right)}{2}$.