= Solution
An <algebraic element> $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its <minimal polynomial>, equivalently $[K(\alpha):K]$. Since $\alpha$ satisfies $T^2-\alpha^2$ over $K(\alpha^2)$, the <tower law> gives $[K(\alpha):K(\alpha^2)]\leq2$ and makes that degree divide the odd integer $[K(\alpha):K]$. It must therefore be one, proving $\boxed{K(\alpha)=K(\alpha^2)}$.
For the addition assertion, let polynomial relations of degrees $r,s$ express $\alpha^r$ and $\beta^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by $\alpha^i\beta^j$ for $0\leq i<r$, $0\leq j<s$ contains one and is closed under multiplication by $\alpha+\beta$. Thus $1,\alpha+\beta,\ldots,(\alpha+\beta)^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so \b[$\alpha+\beta$ is algebraic].
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