Problem 1
First notice $ (x-c^\frac{1}{p})^p = x^p-c $.
So $ F(c^\frac{1}{p}) $ is the splitting field of $ x^p-c $.
Now suppose that the polynomial is reducible in some field $ K \supset F $ in which $ (x-c^\frac{1}{p})^m $ is one of the irreducible factors of $ x^p-c $ for some m < p.
Then $ c^\frac{m}{p} \in K $ but because p is prime, there is an n so that $ nm \equiv 1 $ (mod p) and so $ c^\frac{nm}{p} = c^r c^\frac{1}{p} $. Thus $ c^\frac{1}{p} \in K $ so $ K \supset F(c^\frac{1}{p}) $.
Thus, if $ x^p-c $ is not irreducible over F, $ F \supset F(c^\frac{1}{p}) $ and F contains a root of $ x^p-c $.
