The image is kinda blurry, but from what I can gather, the notation in the attachment means
[tex]p_X[x_k]\equiv\mathbb P(X=k)=\begin{cases}(1-p)^{k-1}p&\text{for }k=1,2,\ldots\\0&\text{otherwise}\end{cases}[/tex]
Then
a) [tex]p_X[x_k\le4]\equiv\mathbb P(X\le4)=\mathbb P(X=1)+\mathbb P(X=2)+\mathbb P(X=3)+\mathbb P(X=4)[/tex]
b) [tex]p_X[1<x_k\le2]\equiv\mathbb P(1<X\le2)=\mathbb P(X=2)[/tex]
c) [tex]p_X[1\le x_k\le2]\equiv\mathbb P(1\le X\le2)=\mathbb P(X=1)+\mathbb P(X=2)[/tex]
