Step 1: Plan the approach.
We solve each inequality on its own, then keep only the values that work for both, and finally count the whole numbers there.
Step 2: Tackle the first inequality.
Look at $x^2-2x+8>0$. Find its discriminant: $D=(-2)^2-4(1)(8)=4-32=-28$.
Step 3: Read the discriminant.
Since $D$ is negative and the leading coefficient is positive, this parabola never touches the x-axis and stays above it. So $x^2-2x+8>0$ holds for every real number. This inequality puts no limit on $x$.
Step 4: Tackle the second inequality.
Now $x^2-3x+2\le 0$. Factor it: $x^2-3x+2=(x-1)(x-2)$. So we need $(x-1)(x-2)\le 0$.
Step 5: Solve the factored form.
A product of two factors is negative or zero between its roots. The roots are $1$ and $2$. So the solution is $1\le x\le 2$.
Step 6: Combine both.
The first allowed all real numbers, so the common region is just $1\le x\le 2$.
Step 7: Count the integers.
The whole numbers in $[1,2]$ are $1$ and $2$, which is $2$ integers.
\[ \boxed{2} \]