Problem 1. follows from the fact that the network weight of a compact Hausdorff space equals the weight of it, but this might be unfamiliar. A dedicated proof of this fact:
Let $\mathcal{T}$ be the original topology on $X$, the compact Hausdorff one. We can assume that $X$ is infinite, otherwise $X$ is discrete and finite, and its set of singletons is the required base.
For every pair $\{x,y\}$ with $x \neq y$ from $X$, we pick $U(x,y)$ and $V(x,y)$ from $\mathcal{T}$ such that $x \in U(x,y), y \in V(x,y), U(x,y) \cap V(x,y) = \emptyset$. Let $\mathcal{T}'$ be the topology generated by all sets $U(x,y)$ and $V(x,y)$, so that $\mathcal{T}'$ has a base of weight at most $|X|$ (as the cardinality of $X$ and the set of pairs from $X$ are equal, $X$ being infinite).
The identity map $f(x) = x$ from $(X,\mathcal{T})$ to $(X,\mathcal{T}')$ is a bijection that is continuous (as $\mathcal{T}' \subset \mathcal{T}$) and goes from a compact space to a Hausdorff one (we basically constructed the latter space to be Hausdorff), and so is a closed map as well, and hence a homeomorphism, from which we conclude that $\mathcal{T} = \mathcal{T}'$ and we are done.