Why can't a left-recursive, non-deterministic, or ambiguous grammar be LL(1)?
I have done some more research, and I think I've found a solution for the 1st and 2nd questions, as for the 3rd one, I found an existing solution on StackOverflow for it, the proof attempts are written below:

We start by writing the three rules of the definition of an LL(1) grammar:

For every production A -&gt; α | β with α ≠ β:


FIRST(α) ∩ FIRST(β) = Ø.
If β =&gt;* ε then FIRST(α) ∩ FOLLOW(A) = Ø (also, if α =&gt;* ε then FIRST(β) ∩ FOLLOW(A) = Ø).
Including ε in rule (1) implies that at most one of α and β can derive ε.


Proposition 1: A non-factored grammar is not LL(1).

Proof:

If a grammar G is non-factored then there exists a production in G of the form:

A -&gt; ωα1 | ωα2 | ... | ωαn


(where αi is the i-th α, not the symbols α and i), with α1 ≠ α2 ≠ ... ≠ αn. We can then easily show that:

∩(i=1,..,n) FIRST(ωαi) ≠ Ø


which contradicts rule (1) of the definition, thus, a non-factored grammar is not LL(1). ∎

Proposition 2: A left-recursive grammar is not LL(1).

Proof:

If a grammar is left-recursive then there exists a production in G of the form:

S -&gt; Sα | β


Three cases arise here:


If FIRST(β) ≠ {ε} then:

    FIRST(β) ⊆ FIRST(S)

=&gt;  FIRST(β) ∩ FIRST(Sα) ≠ Ø

which contradicts rule (1) of the definition.
If FIRST(β) = {ε} then:

2.1. If ε ∈ FIRST(α) then:

ε ∈ FIRST(Sα)

which contradicts rule (3) of the definition.

2.2. If ε ∉ FIRST(α) then:

    FIRST(α) ⊆ FIRST(S) (because β =&gt;* ε)

=&gt;  FIRST(α) ⊆ FIRST(Sα) ........ (I)

we also know that:

FIRST(α) ⊆ FOLLOW(S) ........ (II)

by (I) and (II), we have:

FIRST(Sα) ∩ FOLLOW(S) ≠ Ø

and since β =&gt;* ε, this contradicts rule (2) of the definition.


In every case we arrive at a contradiction, hence, a left-recursive grammar is not LL(1). ∎

Proposition 3: An ambiguous grammar is not LL(1).

Proof:

While the above proofs are mine, this one is not, it's by Kevin A. Naudé which I got from his answer that is linked below:

https://stackoverflow.com/a/18969767/6275103