Does that slight change between Exercises 1 and 2 make a difference to their solutions ? Is it wrong to use recycle Exercise 1's solution for Exercise 2?
Exercise 1: Prove or disprove that $$\forall a{\in}\mathbb Z\; \forall b{\in}\mathbb Z\; \exists c{\in}\mathbb Z\; a|(b+c).\tag1$$ Exercise 2: Prove or disprove that $$\forall a{\in}\mathbb Z\; \exists c{\in}\mathbb Z\; \forall b{\in}\mathbb Z\; a|(b+c).\tag2$$
My solution to Exercise 1:
Let $a_0$ be a fixed, but arbitrary integer.
Let $b_0$ be a fixed, but arbitrary integer.
Choose $c_0 \in \mathbb Z.$
I must prove that $a$ divides $b+c$. I must prove that $\exists k \in \mathbb Z$ such that $a = k(b+c).$
$a = kb+kc.$
$c = \frac{a-kb}{k}.$
Choose $c = \frac{a-kb}{k}$. I have shown that $\forall a \in \mathbb Z, \forall b \in \mathbb Z, \exists c \in \mathbb Z$ such that $a$ divides $b+c.$
It can be useful to try to read it out in English. "Given any number $a$ and a second number $b$, does there exist (AKA can I always find at least one) third number $c$ such that $a|(b+c)$.
Before trying to prove it if it doesn't seem obvious, try it with some examples. For example, what about $a=5$, $b=2$? Can you find a $c$ such that $5|(2+c)$?
There are in fact an infinite number of examples, but there are two that are simplest: Every number divides itself and every number divides 0. Can you generalize either of these to find a $c$ such that either $a=b+c$ or $0=b+c$?
The second claim asks, given a generic number $a$, does there exist a single number $c$ such that for EVERY possible number $b$, $a|(b+c)$.
Again, try an example. Say $a=5$. Now, can you think of number $c$ such that for every $b$, $5|(b+c)$? Hint: If $5|n$ then $5\not| (n+1)$