Relation of two Functions and Logic For two functions, $f \colon \mathbb{R} \to \mathbb{R}$ and $g \colon \mathbb{R} \to \mathbb{R}$, we define the set, $\Omega^g_f=\{x \in \mathbb{R}: f(x)<g(x) \}$. We say that the function $f$ loves $g$ when
$\forall x \in \Omega^g_f, \exists y \in \Omega^f_g $ such that $x<y$.
a) If $f$ loves $g$, and $g$ loves $h$, prove that $f$ loves $h$.
b) For any $t \in \mathbb{R}$, let $f_t(x)=f(x)+t$. For every function $f$ there exist a function $g$ such that, for every $t \in \mathbb{R}$, $g$ loves $f_t$.
I already think a) is not true. I found a counterexample. Take $g(x)=\sin(x)$, $h(x) = 0.5$, $f(x) = -0.5$. For b) I am not sure how to go about it since the question says 'there exists a function' so I have to consider all possible functions and only one has to work. I understand that $f_t$ is just translated up/down by $t$ units.

Statement a) is false and your counterexample is perfect.
Statement b) is true. Let us prove it. Essentially, the proof is based on the fact that the total order on $\mathbb{R}$ has no greatest element.
Let $f \colon \mathbb{R} \to \mathbb{R}$ be a function. Consider the function $g \colon \mathbb{R} \to \mathbb{R}$ defined by $g(x) = f(x) + x$.
We have to show that, for every $t \in \mathbb{R}$, $g$ loves $f_t$, that is, for every $x \in \Omega_{g}^{f_t}$ there exists $y \in \Omega_{f_t}^{g}$ such that $x < y$.
So, let us fix $ t \in \mathbb{R}$. Let $x \in \Omega_{g}^{f_t}$, that is,  $x \in \mathbb{R}$ and $f(x) + x = g(x) < f_t(x) = f(x) + t$.
Thus, $x < t$.
Take $y = t + 1$. Hence, $x < t < y$ and  $f_t(y) = f(y) + t < f(y) + y = g(y)$.
Therefore, $y \in \Omega_{f_t}^{g}$. $\qquad$QED
