The integer closest to $\frac xy$ is $n>1$ if $n-\frac12<\frac xy<n+\frac12$ or equivalently $(2n-1)y<2x<(2n+1)y$ (it doesn't really matter what you decide for the boundary cases or if you take $\le$ instead of $<$).
This describes a triangle with vertices $(0,0)$, $(\frac2{2n-1},1)$ and $(\frac2{2n+1},1)$ and that has area $A_n=\frac12(\frac{2}{2n-1}-\frac2{2n+1})=\frac2{4n^2-1}$.
In the special case $n=0$, we obtain a triangle with vertices $(0,0)$, $(1,0)$, and $(1,\frac12)$ and area $A_0=\frac14$.
The sought probability equals
$$ p =\sum_{k=0}^\infty A_{2k}\approx 0.4646.$$
This sum is also essentially the imaginary part of $\sum_{n=1}^\infty \frac{i^n}n$. More precisely (with the $\frac54$ coming from special treatment for $A_0$ and $\frac in$),
$$ p = \frac54-\Im\left(\sum_{n=1}^\infty \frac{i^n}n\right)=\frac54-\Im(\ln i)=\frac{5-\pi}4.$$
Hint: If $x \equiv 0\pmod{5}$, then $x^4 \equiv 0\pmod{5}$. If $x \not\equiv 0\pmod{5}$, then $x^4 \equiv 1\pmod{5}$ by Fermat's Little Theorem.
Alternatively, if you don't know modular arithmetic, you can do the following to get the same result.
If $x = 5q+r$ for some integers $q$ and $r$ with $0 \le r \le 4$, then we have $x^4 = (5q+r)^4$ $= 625q^4+500q^3r+150q^2r^2+20qr^3+r^4$ $= 5(125q^4+100q^3r+30q^2r^2+4qr^3)+r^4$ where $r^4$ is one of $\{0, 1, 16, 81, 256\}$. Thus, if $x$ is divisible by $5$, then so is $x^4$, and if $x$ is not divisible by $5$, then $x^4$ is one more than a multiple of $5$.
Hence, $x^4-y^4$ is divisible by $5$ iff $x$ and $y$ are both divisible by $5$ or both not divisible by $5$.
Here is how to finish the problem in case you are still stuck:
There are $5n(5n-1)$ total ways to choose $x$ and $y$ without replacement. There are $n(n-1)$ ways to choose $x$ and $y$ without replacement such that both are divisible by $5$, and there are $4n(4n-1)$ ways to choose $x$ and $y$ without replacement such that both are not divisible by $5$. Hence, the probability that $x^4-y^4$ is divisible by $5$ is $\dfrac{n(n-1)+4n(4n-1)}{5n(5n-1)} = \dfrac{17n-5}{25n-5}$.
Best Answer
Hint: to get an even sum, you need two odds or two evens.
Added: to get two evens is $\frac 59 \cdot \frac 48$. Can you get the chance of two odds? As they are disjoint, you can add them.