Let $Y$ be a smooth complex projective curve of genus two, $X$ a Galois cover of degree two of $Y$ and $K$ the canonical divisor of $X$. Let $i$ be the involution of $X$ over $Y$. Can one find a point $P$ on $X$ such that, if $Q=i(P)$, the divisor $5P+3Q$ is linearly equivalent to $2K$?
No. Note that $P\neq Q$ since $i$ is fixed-point free. Since $i^*K=K$, one would have $5P+3Q\equiv 3P+5Q$ (where $\equiv$ means linear equivalence), hence $2P\equiv 2Q$, so $P$ and $Q$ are Weierstrass points on the hyperelliptic curve $Y$. But then $4P\equiv 4Q\equiv K$, and your condition becomes $P-Q\equiv 0$, hence $P=Q$, a contradiction.