List four elements of each of these sets I am currently having trouble understanding these sets.

I am on letter a and I worked out that k is an element of P, where p is composed of positive integers, such that "k-1 is a multiple of 7." No clue what to do with this part.
$\mathbb{P}$ is weird notation for positive integers. (I would read it as the primes). Anyway, you just need to find four positive integers $k$ such that $k-1$ is a multiple of 7. This means $k$ is one more than a multiple of 7. 7 is a multiple of 7, so $k=8$ is in the set. Just think of a few other multiples of 7 and proceed accordingly.
This is my answer:  

a)
$\mathbb{P}$ is a notation used for positive integers (and a lot of times a notation for prime numbers), I would take it as prime numbers. So you need to find $4$ prine numbers $p_i,~i\in\mathbb{N},1\le i\le 4$ such that $7\mid(p_i+1)$ so you need to $7q+1,~q\in\mathbb{N}$ be a prime number.  

Obviously if $q$ is odd then $7q+1$ is even then it is prime if $7q+1=2\Rightarrow7q=1$ and this is not possible because $q\in\mathbb{N}$.  

So $q$ is even lets find $p_i$:
$q=2\Rightarrow7k+1=7\cdot2+1=15$ and it is not prime $(3\mid15)$.
$q=4\Rightarrow7k+1=7\cdot4+1=29$ and it is prime, so $p_1=29$.
$q=6\Rightarrow7k+1=7\cdot6+1=43$ and it is prime, so $p_2$=43.
$q=8\Rightarrow7k+1=7\cdot8+1=57$ and it is not prime $(3\mid57)$.
$q=10\Rightarrow7k+1=7\cdot10+1=71$ and it is not prime $(3\mid57)$.
$q=12\Rightarrow7k+1=7\cdot12+1=85$ and it is not prime $(5\mid85)$.
$q=14\Rightarrow7k+1=7\cdot14+1=99$ and it is not prime $(3\mid99)$.  $q=16\Rightarrow7k+1=7\cdot16+1=113$ and it is prime, so $p_3=113$.  $q=18\Rightarrow7k+1=7\cdot18+1=127$ and it is prime, so $p_4=127$. 

Then this are your four elements: $k\in\{29,43,113,127\}$.  

b)
$x\in\{$apple, strawberry, pear, peach$\}$  

c)
$\displaystyle x\in\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\right\}$  

d)
$n\in\{-1,-2,-3,-4\}$  

e)
Again $\mathbb{P}$, lets take $n\in\{2,3,5,7\}$.