How can the following be false based on the information about continuous functions?

You are correct: statement (2) is always true by the intermediate value theorem, since $\frac{f(0)+f(1)}{2}$ is between $f(0)$ and $f(1)$.

Statement (1) is not true for all $f$. A simple counterexample is the constant function $f(x)=1$.

The first statement reminds of the Mean Value Theorem for Integrals, but… The Mean Value Theorem for Integrals says that if $f$ is continuous on $[a,b]$, then there exists $x\in[a,b]$ such that $$f(x)=\frac{1}{b-a}\int_a^b f(t)\,dt.$$ In this case, $a=-1$ and $b=1$, so we can guarantee that there exists $x\in[-1,1]$ such that $$f(x)=\frac{1}{2}\int_{-1}^1 f(t)\,dt.$$ This is very similar to the given equality, but without the $1/2$. And that suggests that statement (1) is probably false. After realizing that, it's not difficult to come up with a counterexample.

As for statement (2), it's actually true by the Intermediate Value Theorem, since $f$ is continuous.