No. For a very simple example, take $\Omega = \{a,b,c\}$ consisting of three points, with $\mathcal{F} = 2^\Omega$ and $P(A) = |A|/3$ the uniform measure. Let $\mathcal{G} = \{\{a\}, \{b,c\}, \Omega, \emptyset\}$. Then $\mathcal{G}$ is a proper sub-$\sigma$-algebra but no nontrivial event in $\Omega$ is independent of it.

No. For a very simple example, take $\Omega = \{a,b,c\}$ consisting of three points, with $\mathcal{F} = 2^\Omega$ and $P(A) = |A|/3$ the uniform measure. Let $\mathcal{G} = \{\{a\}, \{b,c\}, \Omega, \emptyset\}$. Then $\mathcal{G}$ is a proper sub-$\sigma$-algebra but no nontrivial event in $\mathcal{F}$ is independent of it.

Edit: Andres Caicedo asks for a non-atomic example. George Lowther gave one. Another is supplied by taking $\Omega = [0,1]$, $\mathcal{G} = \mathcal{B}_{[0,1]}$ the Borel $\sigma$-field, $\mathcal{F} = \mathcal{L}$ the Lebesgue $\sigma$-field which is the completion of $\mathcal{G}$, and $P = $ Lebesgue measure. Now by definition of $\mathcal{L}$, for any $A \in \mathcal{F}$ we have $A = B \cup N$ where $B \in \mathcal{G}$ and $P(N) = 0$. Then $P(A \cap B) = P(B) = P(A)$ so $A$ and $B$ are independent iff $P(A) = 0$ or $1$.