= Solution
The space of <real numerical divisor classes> is $N^1(X)_{\mathbb R}$. Write any big class as $[D]=[A]+[E]$, with $A$ an <ample real divisor> and $E$ effective. For a sufficiently small perturbation $\delta\in N^1(X)_{\mathbb R}$, the class $[A]+\delta$ remains in the open <ample cone>, so $[D]+\delta$ remains big by part (v). Thus $\operatorname{Big}(X)$ is open. Positive scaling and addition preserve the ample-plus-effective expression, so it is a <convex cone>.
For $n\geq1$, fix a very ample divisor $H$. The linear functional $\ell([D])=D\cdot H^{n-1}$ is strictly positive on every big class: the ample part contributes positively and the effective part nonnegatively. No nonzero linear subspace can be contained in this cone, since it would contain both $v$ and $-v$. In particular the zero class is not big in positive dimension. When $n=0$, $N^1(X)_{\mathbb R}=0$, so there is no positive-dimensional subspace to consider. Therefore
$$
\boxed{\operatorname{Big}(X)\text{ is an open convex cone and contains no positive-dimensional vector subspace}.}
$$
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