Do i need to find the elements of all powerset to answer such questions? Let $M = \{\{2\},2,5,7,0,\{23\}\}$ .
Q1. $\{7, \{23\}\} \in \mathcal{P}(M)$?
A1. Since they are elements in m then they are true.
Q2. $\{\{2\},\{5,7\}\} \subseteq \mathcal{P}(M)$? However here I am confused? What should I do?
A2. To solve it I usually turn the symbol back to $\in$, and then if the answer was true I just flip it to false? is this approach right?
 A: You should turn to the definition of $\subseteq$: the statement that $A\subseteq B$ is equivalent to saying that, for every $x\in A$ it is also the case that $x\in B$. Since you know that the statement $S\in P(M)$ is true if and only if $S\subseteq M$ - as you use to solve the first question - to check if a given set $S$ is a subset of $P(M)$, you look at every element within $S$ and ask if that is a subset of $M$. In particular, you can verify that $\{2\}$ and $\{5,7\}$ are both elements of $P(M)$, thus $\{\{2\},\{5,7\}\}$ is a subset of $P(M)$.
Note that this is different from your suggestion of changing $\subseteq$ to $\in$ and reversing the answer - this doesn't work; you can find sets which are simultaneously subsets and elements of $P(M)$ or (somewhat easier) sets which are neither a subset nor an element of $P(M)$.
A: Your approach isn't necessarily right, particularly for transitive sets, which are sets in which every element is also a subset of the set.  In this example, for instance, note that $\{2\} \in M$ and also $\{2\} \subseteq M$ (because $2 \in M$).
For Question $2$, whether $\{\{2\}, \{5, 7\}\} \subseteq \mathscr P(M)$ is the same question as whether $\{2\} \in \mathscr P(M)$ and also $\{5, 7\} \in \mathscr P(M)$.  Can you see whether that's true?
A: You assume the following principle: 
       if an object ( here, a set)  belongs to a set S , it is not included in S.  

Now, let's consider the number 2 that is usually defined as : 
          2 = {0,1} = {  {  } ,  {0}  } . 

One can see that 
(1) the number 1, that is : {0} ,  is an element of 2 , it belongs to the number 2 
(2) the number 1 is also a subset of 2 ( all the elements of {0} are also elements of 2, since {0} has only one element, namely the number 0 , and 0 is an element of 2). 
As a general rule: every natural number is both a member and a subset of its successor. 
