{$ \rightarrow, \wedge $} and Quantifiers 
*

*Implication goes with $ \forall $

*Conjunction goes with $ \exists $
Does the first statement mean that a conjunction should not appear in the scope of a $ \forall $, or does it mean that a conjunction can appear as long as it is not the main connective?
e.g. $ \forall x (P(x) \wedge Q(x) \rightarrow R(x))$
Same question for the second statement
 A: No, there are no rules that forbide the use of $\forall$ with conjunction.
$\forall x (A(x) \land B(x))$ is a correct formula.
Maybe you are alluding to the so-called restricted (or: bounded) quantifiers:

$\forall x \ \in A : P(x)$ and $\exists x \ \in A :P(x)$

that are abbreviations respectively for the two formulae :

$\forall x (x \in A \to P(x))$ and $\exists x (x \in A \land P(x))$.

A: Neither. What is meant is that the "all" sentences you typically want to express translate as $\to$, and typical "there is" sentences translate as $\land$.
It is perfectly legal to write
$$\forall x (Swan(x) \land White(x))$$
-- if what you want to say is that all that your model contains is white swans.
Likewise,
$$\exists x(Swan(x) \to White(x))$$
is a perfectly grammatical sentence -- which becomes true as soon as there is anything which is not a swan, or anything which is white. It's just that this is not something you will very often find yourself in the need to say.
What is meant by these rules of thumbs that when expressing an "all" statement, what you mean is usually $\forall x (\phi(x) \to \psi(x))$, and sentences with "there is" usually mean what is formalized as $\exists x (\phi(x) \land \psi(x))$. If you are sure it's something else you want to express, then as far as the language of predicate logic is concerned, you are allowed to do so: $\forall$ and $\exists$ can range over arbitrary formulas with arbitrary connectives.

Also note that the two statements left and right of $\to$ resp. $\land$ can be "compound" expressions:
For instance, adjective + noun or noun + relative clause translate as $\land$, and predications with "or" may also occur, so e.g.
$$\text{Every old swan is white}\\
= \forall x ((Swan(x) \land Old(x)) \to White(x))$$
$$\text{Every swan is an animal which is white}\\
= \forall x (Swan(x) \to ((Animal(x) \land White(x)))$$
$$\text{Every swan is white or gray}\\
= \forall x (Swan(x) \to ((White(x) \lor Gray(x)))$$
and the same for existential expressions.
A: The $\forall$ usually goes with $\to$, and the $\exists$ usually goes with $\land$, but this is not a hard rule: it's perfectly ok to have any kind of operator inside any quantifier, even as the main operator ... it's just that many sentences we use in real life, when translated to logic, end up with the pairings you indicated.
