(1) The corresponding English meaning: If $P(x)$ is true for all $x$, or if $Q(x)$ is true for all $x$, then for all $x$, either $P(x)$ is true or $Q(x)$ is true. This is always true and hence valid. To understand deeply, consider $X = \{3,6,9,12\}$. For LHS of implication to be true, either $P(x)$ must be true for all elements in $X$ or $Q(x)$ must be true for all elements in $X$. In either case, if we take each element $x$ in $X$, either one of $P(x)$ or $Q(x)$ will be true. Hence, this implication is always valid.
(If still in doubt, let $P(x)$ mean $x$ is a multiple of $3$ and $Q(x)$ means $x$ is a multiple of $2$)
(2) The corresponding English meaning: If $P(x)$ is true for at least one $x$, and if $Q(x)$ is true for at least one $x$, then there is at least one $x$ for which both $P(x)$ and $Q(x)$ are true. This is not always true as $P(x)$ can be true for one $x$ and $Q(x)$ can be true for some other $x$ . To understand deeply, consider $X = \{3,6,9,12\}$. Let $P(x)$ be $x$ is a multiple of $9$ and $Q(x)$ be $x$ is a multiple of $6$. Now, LHS of implication is true, since $P(x)$ is true for $x = 9$, and $Q(x)$ is true for $x = 6$. But RHS of implication is not true as there is no $x$ for which both $P(x)$ and $Q(x)$ holds. Hence, this implication is not valid.
(3) If for each $x$, either $P(x)$ is true or $Q(x)$ is true then $P(x)$ is true for all $x$ or $Q(x)$ is true for all $x$. Just one read is enough to see this is an invalid implication. Consider set {2,4,5}. Here every element is either a multiple or 2 or 5. But all elements are neither multiple of 2 nor 5.
(4)If there is at least one $x$ for which either $P(x)$ or $Q(x)$ is true then either it is not the case that $P(x)$ is true for all $x$ or $Q(x)$ is true for at least one $x$. This is clearly invalid as LHS of implication becomes true if $P(x)$ is true for some $x$ and $Q(x)$ is not true for any $x$, but RHS will be false (if $P(x)$ is true for all $x$).
A little modification to the statement is enough to make it valid:
$$\exists (x) (P(x) \vee Q(x)) \implies \sim (\forall (x) \sim P(x)) \vee \exists (x) Q(x)$$
which means if there is at least one $x$ for which either $P(x)$ or $Q(x)$ is true then
either it is not the case that $\sim P(x)$ is true for all $x$ (which means P(x) is true for some $x$) or $Q(x)$ is true for some $x$.
Note
De Morgan's law is applicable in first order logic and is quite useful:
$$\forall(x)(P(x)) \equiv \neg \exists (x)(\neg P(x))$$
This is a logical reasoning statement which means if $P(x)$ is true for all $x$, then there can never exist an $x$ for which $P(x)$ is not true. This formula is quite useful in proving validity of many statements as is its converse given below:
$$\exists(x)(P(x)) \equiv \neg \forall (x)(\neg P(x))$$