Employing the creative microscoping method and the Chinese remainder theorem for polynomials, we give a parametric q-supercongruence and two Dwork-type q-supercongruences. As a corollary we deduce that, for primes \(p\equiv 1\pmod {4}\) and \(r\geqslant 1\) , \( \sum _{k=0}^{p^r-1}(6k+1)\frac{(\frac{1}{2})_k(\frac{1}{4})_k^2}{k!^3}4^k \equiv p \sum _{k=0}^{p^{r-1}-1}(6k+1)\frac{(\frac{1}{2})_k(\frac{1}{4})_k^2}{k!^3}4^k \pmod {p^{3r}}, \) and a similar result for primes \(p\equiv 3\pmod {4}\) and \(r\geqslant 2\) , \( \sum _{k=0}^{p^r-1}(6k+1)\frac{(\frac{1}{2})_k(\frac{1}{4})_k^2 4^k }{k!^3} \equiv p^2 \sum _{k=0}^{p^{r-2}-1}(6k+1)\frac{(\frac{1}{2})_k(\frac{1}{4})_k^2 4^k }{k!^3} \pmod {p^{3r-2}}, \) where \((x)_0=1\) and \((x)_k=x(x+1)\cdots (x+k-1)\) for \(k\geqslant 1\) .
