= Solution
A <pointed cone> has no nonzero line through the origin: equivalently $K\cap(-K)=\{0\}$. If $A\in K\cap(-K)$, its <quadratic form> vanishes on the <nonnegative orthant>. Testing the coordinate <vectors> gives $A_{ii}=0$. Testing $e_i+e_j$ then gives
$$
0=(e_i+e_j)^TA(e_i+e_j)=2A_{ij}.
$$
Since $A$ is a <symmetric matrix>, every entry is zero. Thus $\boxed{K\cap(-K)=\{0\}}$.
The <identity matrix> gives an <interior> point. For a symmetric perturbation $E$ with <operator norm> $\|E\|_2<1$, the <Cauchy-Schwarz inequality> gives
$$
x^T(I+E)x\geq(1-\|E\|_2)\|x\|_2^2>0\quad(x\ne0).
$$
This open <norm> ball around $I$ lies even in the <positive semidefinite cone>, hence in $K$. Consequently $\boxed{I\in\operatorname{int}K}$ and the <interior> is nonempty.
More generally, the <interior of the copositive cone> consists exactly of <strictly copositive matrices>. Positivity on the <compact> nonnegative <unit sphere> has a positive minimum and persists under small perturbations; a zero there is destroyed by an arbitrarily small negative multiple of $I$.
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