Existence of a non-null and non-negative vector in $F\cup F^\perp$

Yes. For any two closed convex cones $C,D$ we have $(C+ D)^\circ=C^\circ\cap D^\circ$, where $C^\circ$ is the dual cone: $C^\circ=\{y\in\mathbf{R}^n:\forall c\in C:\langle y,c\rangle\ge 0\}$. A standard duality theorem is $C^{\circ\circ}=C$ for every closed convex cone $C$. Hence, the dual equality holds: $(C\cap D)^\circ=C^\circ+ D^\circ$.

Let $P$ be the cone of non-negative vectors. Apply this to $P$ and $F$. Assume that $P\cap F=\{0\}$. Then $(P\cap F)^\circ =\mathbf{R}^n$. Since $P^\circ=P$ and $F^\circ=F^\bot$, this yields $P+F^\bot=\mathbf{R}^n$. Fix any nonzero $\xi\in P$ (this requires $n\ge 1$ which I discretely added to the assumptions). Then $-\xi=p+\eta$ for some $p\in P$ and $\eta\in F^\bot$. Hence $-\eta=p+\xi$ is a nonzero element in $P\cap F^\bot$.

Yes. This follows from Farkas' lemma. Let $\vec{v}_1$, $\vec{v}_2$, ..., $\vec{v}_k$ be a basis of $F$ and let $\vec{e}_1$, $\vec{e}_2$, ..., $\vec{e}_n$ be the standard basis of $\mathbb{R}^n$. Suppose that there is no nonzero nonnegative vector in $F^{\perp}$. In other words, suppose there is no nonzero solution to the linear inequalities: $$\vec{v}_i \cdot \vec{x} = 0, \ 1 \leq i \leq k \qquad \qquad \vec{e}_j \cdot \vec{x} \geq 0, \ 1 \leq j \leq n.$$

Then Farkas' lemma tells us that there must be some linear relation $$\sum a_i \vec{v}_i + \sum b_j \vec{e}_j = 0$$ with nonnegative coefficients (and not all coefficients zero). Then $- \sum a_i \vec{v}_i$ is an element of $F$ with nonnegative entries. Moreover, we can't have $- \sum a_i \vec{v}_i = 0$, as the $\vec{v}_i$ are linearly independent and the $\vec{e}_j$ are as well.