May I have assistance on the following two exam questions where $G$ is finite group and $p$ is prime. $\DeclareMathOperator{\Syl}{Syl}$
1)$\;\;$ If $P\in \Syl_{p}(G)$ and $K$ is a subgroup of $G$ containing $P$, prove that $P\in \Syl_{p}(K)$; and$\;\;$ 2)$\;\;$ If $K$ is a subgroup of $G$ and $Q\in \Syl_{p}(K)$, prove that $gQg^{-1}\in \Syl_{p}(gKg^{-1})$.
My attempt for 1):$\;\;$ Since $P\in \Syl_{p}(G)$, then $G$ must consist of a number of Sylow $p$-subgroups, say $G= \{P_{1}, P_{2}, ...P, ...P_{n}\}$, at least one of which, say $P$, is contained in $K$.
Hence one can write that $\Syl_{p}(K)\in \Syl_{p}(G)$ and thus that $P\in \Syl_{p}(K)$.
My attempt for 2) follows similar outline. If $Q\in \Syl_{p}(K)$, this implies that $K$ can be written as a set of Sylow $p$-subgroups, say $K=\{Q_{1},Q_{2},..Q,..Q_{n}\}$.
Hence $gKg^{-1}=\{gQ_{1}g^{-1},...gQg^{-1}, ..gQ_{n}g^{-1}\}$, and so we can state that $gQg^{-1}\in \Syl_{p}(gKg^{-1})$.
I am new to abstract algebra and am finding it very difficult. I do not know if my attempts carry any validity.
Thanks in advance for any "wisdom".
Your attempts are incomplete, you don't argue why "one can write that $\operatorname{Syl}_{p}(K)\in \operatorname{Syl}_{p}(G)$ and thus that $P\in \operatorname{Syl}_{p}(K)$".
The correct argument relies on the definition of Sylow $p$-subgroups of a group. $P\subset G$ is called a Sylow $p$-subgroup of $G$ if order of $P$ is the largest power of $p$ dividing order of $G$.
Let order of G be $p^mr,$ where $p\not| r$. Then order of $P$ is $p^m$ and it is the largest exponent of $p$ dividing order of $K$ (which follows from Lagrange's theorem), hence part (1) follows.
For part (2), observe that $gQg^{-1}$ is a subgroup of $gKg^{-1}$ and order of $gQg^{-1}$ is same as order of $Q$ and order of $gKg^{-1}$ is same as order of $K$. Now, we know that order of $Q$ is the largest power of $p$ dividing order of $K$, thus order of $gQg^{-1}$ is the largest power of $p$ dividing order of $gKg^{-1}$, and we are done.