If $\gcd(a,b)=1$ and $p \mid a$ then $p \nmid b$. But how one can show that $\gcd(p^k,b)=1$? And can one show that if $p^k \nmid b$, then $\gcd(p^k, b)=1$? And can one show that if $\gcd(p^k,b)=1$, then $p^k \nmid b$?
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Since $\gcd(p^k,b)=1$ doesn't hold in general, I'll assume that you intended the first question to share the assumptions of the first sentence, $\gcd(a,b)=1$ and $p\mid a$. In that case, since $p\nmid b$ and $p$ is the only prime factor in $p^k$, it follows that $p^k$ and $b$ have no factors in common, and thus $\gcd(p^k,b)=1$.
The answer to the second question is no: $p^2\nmid p$, but $\gcd(p^2,p)\ne1$.
The answer to the third question is yes, for if $p^k\mid b$, then $p^k\mid\gcd(p^k,b)$, so $\gcd(p^k,b)\ne1$.