(a) If $p$ is a prime other than 2 or 3, then $[R_{12}(Z_p):Z_p]\leq2$, $\Phi_{12}$ is reducible over $Z_p$, and $Z_p$ contains 12th root of unity other than 1.
(b) The field $Z_3$ contains no 11th root of unity other than 1, but $\Phi_{11}$ is reducible over $Z_3$.
I know that if K be a field whose characteristic is either zero or not a divisor of $n$ (a) The degree of every prime factor in $K[x]$ of $\Phi_n$ is $[R_n(K):K]$ (b) If $K$ has $q$ elements, then $[R_n(K):K]$ is the smallest of the strictly positive integers $m$ such that $n|q^m-1$ (c) If K has $q$ elements, then $\Phi_n$ is irreducible over $K$ if and only if the order of the coset $q+(n)$ in $Z_n$ is $\varphi(n)$
Please help me
One efficient way to analyse problems of this type is to look for the smallest (finite) field extension of your ground field that contains a root. This can tell you the degree of the irreducible factors of your polynomial.
For your question (b), for instance, you’re asking about eleventh roots of unity in characteristic three. What’s the smallest field, cardinality $3^m$, so that its multiplicative group, order $3^m-1$, has a subgroup of order $11$? In other words, what’s the smallest $m$ with $11|(3^m-1)$? You can do this by inspection, knowing that the successive powers of $3$ are $3$, $9$, $27$, $81$, $243$ — whoops, there you are, at $3^5$, so that $\Phi_{11}$ has quintic $\Bbb Z_3$-irreducible factors, two of them.
I leave it to you to use these ideas for part (a).