If is true f(n) = Θ(g(n)) and if f(n) = o(h(n)) then g(n) = o(h(n))?
From $f(n) = \Theta(g(n))$ you know that, for some positive constants $c \in \mathbb{R}^+$ and $\eta \in \mathbb{N}$, and for all $n \ge \eta$, it holds that $f(n) \ge c g(n)$, i.e., $g(n) \le \frac{1}{c} f(n)$.$^1$
Moreover, from $f(n) = o(h(n))$, you know that for any constant $c' &gt; 0$ there is some $\eta'_{c'}$ such that $f(n) \le c' h(n)$ for any $n \ge \eta'_{c'}$.
Pick any constant $c'' &gt; 0$ and let $\eta'' = \max\{\eta, \eta'_{c'}\}$ where $c' = c \cdot c'' &gt; 0$.
From the previous inequalities we have that, for any $n \ge \eta''$:
$$
g(n) \le \frac{1}{c} f(n) \le \frac{1}{c} \cdot  c' h(n) =  c'' h(n),
$$
thus proving that $g(n) = o(h(n))$.

$^1$: In fact, the weaker condition $f(n) = \Omega(g(n))$ suffices.