Are these two ways of completing this inequality both correct: (First one is from proffesor, and second one I did, the answer is at the end correct, but I cant figure out if my number of solutions matches because of some wild luck or is that also an correct way for the solution) First: $$ \begin{aligned} & \left|\frac{2-5 x}{2 x+1}\right| \leq 2, D . P .: 2 x+1 \neq 0 \Rightarrow x \neq-\frac{1}{2} \\ & \frac{|2-5 x|}{|2 x+1|} \leq 2,|2-5 x|=\left\{\begin{array}{c} 2-5 x, 2-5 x \geq 0, x \leq \frac{2}{5} \\ -(2-5 x), 2-5 x<0, x>\frac{2}{5} \end{array}\right\},|2 x+1|=\left\{\begin{array}{c} 2 x+1,2 x+1>0, x>-\frac{1}{2} \\ -(2 x+1), 2 x+1<0, x<-\frac{1}{2} \end{array}\right\} \\ & I: x \in\left(-\infty,-\frac{1}{2}\right) \\ & \frac{2-5 x}{-(2 x+1)}-2 \leq 0 \Rightarrow \frac{5 x-2}{2 x+1}-2 \leq 0 \Rightarrow \frac{x-4}{2 x+1} \leq 0 \Rightarrow x \in\left(-\frac{1}{2}, 4\right] \cap I \Rightarrow x \in\{\varnothing\} \text {. } \\ & \end{aligned} $$ No integer solutions $$ \begin{aligned} & I I: x \in\left(-\frac{1}{2}, \frac{2}{5}\right] \\ & \frac{2-5 x}{2 x+1}-2 \leq 0 \Rightarrow \frac{5 x-2}{2 x+1}+2 \geq 0 \Rightarrow \frac{9 x}{2 x+1} \geq 0 \Rightarrow x \in\left(-\infty,-\frac{1}{2}\right) \cup[0,+\infty) \cap I I \Rightarrow x \in\left[0, \frac{2}{5}\right) . \end{aligned} $$ Integer solution $: x_1=0$. $$ \begin{aligned} & I I I: x \in\left(\frac{2}{5},+\infty\right) \\ & \frac{-(2-5 x)}{2 x+1}-2 \leq 0 \Rightarrow \frac{5 x-2}{2 x+1}-2 \leq 0 \Rightarrow \frac{x-4}{2 x+1} \leq 0 \Rightarrow x \in\left(-\frac{1}{2}, 4\right] \cap I I \Rightarrow \Rightarrow x \in\left(\frac{2}{5}, 4\right] . \end{aligned} $$ Integer solutions: $x_2=1, x_3=2, x_4=3, x_5=4$. Number of integer solutions is 5.
Second:
Image(in final answer there is a typo, it is not (0,4] but [0, 4]): 
Your solution is correct (After fixing the final step to $ x \in [ 0, 4 ] $).
Note that in the first solution, for case II, the conclusion should be $ x \in [ 0, \frac{2}{5} \color{red}{]}$. I'm guessing this is just a typo.
With that, both solution ranges match, are equal to $ x \in [ 0, 4 ]$, which give the same integer solutions.
Side note: My favorite way of dealing with these types of absolute value inequalities (single term only) is just to square and multiply out the denominator (excluding when it is 0) to obtain $ (2 - 5x) ^2 \leq 2^2 (2x+1)^2$, which we can easily shift over and factorize to $(2 - 5x - 2(2x+1) ) ( 2 - 5x+2(2x+1) ) \leq 0 $, and hence arrive at $ x \in [ 0, 4 ]$. It is quick, easy to follow, and avoids a lot of potential computational errors.