Circle1 with $(1,1)$ and $r=1$
Circle2 with $(3,2.5)$ and $ r=2$
Best way to calculate the intersection without a calculator on a piece of paper, I tried many ways which I saw on the internet and spend much time, but my results are mostly wrong.
I really don't know if it's because of the formulas or because my little errors, does someone know a good way? Which is not so complicated? A proper formula, maybe someone can show me how he calculated it?
Now I see that you've changed the question since I posted this answer. The answer below is correct for the question as initially stated.
The same technique will find the points of intersection to the later modified problem.
If you just draw the picture you can see that they don't intersect.But one can also look at their two equations: $$ (x-1)^2 + (y-3)^2 = 1^2 = 1 \tag 1 $$ $$ (x-1)^2 + (y-2.5)^2 = 2^2 = 4 \tag 2 $$ Subtracting the left side of $(1)$ from the left side of $(2)$ and likewise on the right sides, we get $$ (y-2.5)^2 - (y-3)^2 = 4-1 = 3. $$ Expanding the two squares, we get $$ (y^2 - 5y + 6.25) - (y^2 - 6y + 9) = 3. $$ Then: $$ y - 2.75 = 3 \quad\text{so}\quad y = 5.75. $$ Plugging $5.75$ in place of $y$ in $(1)$ we get $$ (x-1)^2 + (5.75-3)^2 = 1 $$ or $$ (x-1)^2 = 1 - 2.75^2. $$ So a square equals a negative number. Hence there is no solution. The two circles do not intersect.