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I think (1) is straightforward: suppose $C$ is a cyclic code of length $n$ with generator polynomial $f(x) \in F_q[x]$. Let $C'$ be the code with generator polynomial $g(x) = (x^n-1)/f(x)$. As I understand the definition of duality for cyclic codes (following page 84 of van Lint, Introduction to Coding Theory, 3rd edition), the dual of $C$ is $C'$. Hence, if $C$ is self-dual then $f(x) = g(x)$ and so $x^n-1 = f(x)^2$. This is impossible when $q$ is odd.

I can't help with (2), except to say that a quick Google search found several papers that look relevant, e.g. Self-dual cyclic codes by N.J.A Sloane and J.G. Thompson and On the Minimal Distance of Binary Self-Dual Cyclic Codes by Bas Heijne and Jaap Top.

Edit: as Zahra's comment says, with the other definition of duality for cyclic codes, the generator polynomial for $C^\bot$ is a multiple of $h(x) = g(x^{-1})x^{\deg g}$. The roots of $h(x)$ are the reciprocals of the roots of $g(x)$, counted with multiplicities. Suppose that $1$ is a root of $x^n-1$ with multiplicity $m$. If $1$ has multiplicity $r$ as a root of $f(x)$, then $1$ has multiplicity $m-r$ as a root of $g(x)$ and $h(x)$. Hence, if $C$ is self-dual then $f(x) = h(x)$, and $m-r = r$. Therefore $m$ is even, and again this is impossible when $q$ is odd.

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I think (1) is straightforward: suppose $C$ is a cyclic code of length $n$ with generator polynomial $f(x) \in F_q[x]$. Let $C'$ be the code with generator polynomial $g(x) = (x^n-1)/f(x)$. As I understand the definition of duality for cyclic codes (following page 84 of van Lint, Introduction to Coding Theory, 3rd edition), the dual of $C$ is $C'$. Hence, if $C$ is self-dual then $f(x) = g(x)$ and so $x^n-1 = f(x)^2$. This is impossible when $q$ is odd.

I can't help with (2), except to say that a quick Google search found several papers that look relevant, e.g. Self-dual cyclic codes by N.J.A Sloane and J.G. Thompson and On the Minimal Distance of Binary Self-Dual Cyclic Codes by Bas Heijne and Jaap Top.