Real-analytic function of two complex variables, holomorphic in first and anti-holo in second, which vanishes on the diagonal is identically zero.

Let's start with part 2). It is indeed redundant to explicitly demand that $f$ be real-analytic. That is however a nontrivial theorem. Hartogs proved that if a function (of several complex variables) is separately holomorphic, then the function is holomorphic. Here, applying that theorem to $g\colon (z,w) \mapsto f(z,\overline{w})$ yields the holomorphicity of $g$ and then the real-analyticity of $f$, as a composition of real-analytic functions.

For part 1), we also use the function $g$. The premise that $f$ vanish on the diagonal means that $g$ vanishes on the conjugate-diagonal $\{(z,\overline{z}) : z \in \mathbb{C}\}$. A modicum of several-variable theory tells us that if $h$ is a holomorphic function on an open subset $U$ of $\mathbb{C}^n$, and there is a $z_0 \in U$ with $h(z_0) = 0$ and $\operatorname{grad} h(z_0) \neq 0$, then the zero set of $h$ is in a neighbourhood of $z_0$ a complex submanifold of dimension $n-1$. Since the conjugate-diagonal is nowhere a complex submanifold of $\mathbb{C}^2$, it follows that $\partial_k g$ vanishes identically there. Iterating the reasoning, it follows that all partial derivatives of $g$ vanish on the conjugate-diagonal, and hence the Taylor series of $g$ with centre $0$ is the zero series, i.e. $g \equiv 0$.