FALSE: an ambiguous CFG always generates an inherently ambiguous language

The displayed grammar is ambiguous: the word $a+a\ast a$ has one parse corresponding to $a+(a\ast a)$ and another corresponding to $(a+a)\ast a$. So the grammar is an ambiguous CFG.

The standard precedence grammar $E\to E+T\mid T,T\to T\ast F\mid F,F\to (E)\mid a\mid b$ generates the same terminal strings. One inclusion follows by simultaneous induction on its $E$-, $T$-, and $F$-derivations, since every displayed production can be simulated by the first grammar. For the reverse inclusion, scan a generated word at parenthesis depth zero: if there is a +, split at the rightmost such +; otherwise, if there is a *, split at the rightmost such *; the remaining factor is either $a$, $b$, or a parenthesized expression. Induction on the word length gives a derivation in the precedence grammar. These rightmost top-level splits are forced by its left-recursive $E$- and $T$-rules, while the $F$-case is forced by the outer symbols, so the resulting parse is unique. In particular, $a+a\ast a$ has only the parse $a+(a\ast a)$. Thus the language has an unambiguous grammar.

By [L1], a language with an unambiguous grammar is not inherently ambiguous. So the ambiguous grammar from step 1.1 does not force its language to be inherently ambiguous, and [A1] is false.