As I commented, the proof of Normal Basis Theorem is not very hard. Please comment me any questions regarding the proof.
As for the exercise, the "enough roots of unity" condition simplifies the proof of this problem. In this proof, we avoid using the Normal Basis Theorem.
Let $n=|G|$ and assume that $F$ contains a primitive $n$-th root of unity (that should be a more precise statement about having enough roots of unity). Regarding $g\in G$ as a $F$-linear endomorphism on $K$. Since $g^n=1$, the minimal polynomial of $g$ divides $X^n-1 \in F[X]$. It is easy to see that if $g=1$ then $\chi(1)=|G|=n$.
So, we assume $g\neq 1$. Let $1<d|n$ be the order of $g$ in $G$. Then $g^d=1$ and any eigenvalue of $g$ must be a $d$-th root of unity. Since distinct field homomorphisms $1,g,\ldots, g^{d-1}$ are linearly independent over $F$, it follows that $X^d-1\in F[X]$ is the minimal polynomial of $g$. Since $F$ contains $n$-th roots of unity, we may assume that $g$ is diagonalizable over $F$. Moreover, the diagonal matrix $D_g$ corresponding to $g$ has all $d$-th roots of unity appearing on its diagonal entries.
By Galois theory, the field $K^g=\{x\in K| gx=x\}$ is a subfield of $K$ with extension degree $n/d$ over $F$. Then there is a basis $\{y_1,\ldots, y_{n/d}\}$ of $K^g$ over $F$. For each $d$-th root of unity $\zeta$, take a eigenvector $x\in K$ of $g$ so that $g x= \zeta x$. Then $\{xy_1, \ldots, xy_{n/d}\}$ forms a set of $n/d$ linearly independent eigenvectors. Thus, each $d$-th root of unity appears on exactly $n/d$ diagonal entries of $D_g$, and it follows that $(X^d-1)^{n/d}$ is the characteristic polynomial of $g$. Since the coefficient of $X^{n-1}$ in $(X^d-1)^{n/d}$ is zero, we have $\mathrm{tr}(g)=0$.
Hence, the character of the representation is the regular representation.
