Claim 1: If $f$ is a pdf of a random variable $X$, then $0 \leq f(x) \leq 1 \quad \forall x\in\mathbb{R}$
Why is this False? Is it because $f(x)$ does not need to be defined on the whole real line?
Claim 2: For two events, A, B, we have $\mathbb{P(A\cup B) = \mathbb{P(A)}} + \mathbb{P(B)}$ iff A, B, are disjoint.
Why is this False? I know that if they are disjoint then the equation is true. So the equation somehow cannot imply mutual exlusivity somehow, right?
for claim 1 consider $f(x) = \begin{cases} 2, & \text{if $0 \lt x \lt \frac 12$} \\ 0, & \text{otherwise} \end{cases}$
This is a perfectly good pdf (try drawing it). What the question is meant to illustrate is that we don't need the value of the pdf to lie in a certain range, we just need the integral over the range of possible values to be exactly 1. As we see above even if the function is bigger than one somewhere, because it's only 'big' over a short range, the integral itself is still equal to 1.
Regarding your comment, assuming that by 'defined' you mean nonzero then perhaps consider a normal distribution. If you reduce the variance you can make it 'taller than 1' at the mean, but it will still integrate to 1 and be nonzero everywhere.
'It seems like the integral should be 'bigger' than the function, but this is not true'
As far as claim 2 goes, I am not completely sure, but it could be that if A and B are not disjoint, but the probability of any outcome in $A \cap B$ is zero then we get $P(A \cap B) = 0$ and the equation $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ would still become $P(A \cup B)=P(A)+P(B)$