1. Given quadratic

   $2x^2-8x+k=0$

   Let its roots be $\alpha$ and $\beta$ with $\alpha\in(1,2),\qquad \beta\in(2,3).$

2. Use sum and product of roots

   For the equation $2x^2-8x+k=0$,

   \qquad \alpha\beta=\frac{k}{2}.$$

3. Use the interval condition with the sum

   Since $\alpha\in(1,2)$, write $1<\alpha<2.$ Since $\beta=4-\alpha$, $\beta=4-\alpha.$

   Now check whether $\beta$ lies in $(2,3)$:

   • If $1<\alpha<2$, then $2<4-\alpha<3.$ So the second condition is automatically satisfied.

   Hence we only need $1<\alpha<2,$ where $\alpha$ is a root and the other root is $4-\alpha$.

4. Express roots explicitly

   Solve the quadratic: $2x^2-8x+k=0$ $x^2-4x+\frac{k}{2}=0$

   Therefore, $x=\frac{4\pm\sqrt{16-2k}}{2}=2\pm\frac{\sqrt{16-2k}}{2}.$

   So the two roots are

   \qquad 2+\frac{\sqrt{16-2k}}{2}.$$ For one root to be in $(1,2)$ and the other in $(2,3)$, we need $$1<2-\frac{\sqrt{16-2k}}{2}<2$$ and automatically the other will be in $(2,3)$ by symmetry.

5. Solve the inequality

   From $1<2-\frac{\sqrt{16-2k}}{2},$ we get $-1< -\frac{\sqrt{16-2k}}{2}$ $1>\frac{\sqrt{16-2k}}{2}$ $2>\sqrt{16-2k}.$

   Also, for the smaller root to be less than $2$, we need $\sqrt{16-2k}>0,$ so the roots are distinct and not both equal to $2$.

   Thus, $0<\sqrt{16-2k}<2.$

   Squaring, $0<16-2k<4.$

   This gives $0<16-2k \Rightarrow k<8,$ and $16-2k<4 \Rightarrow -2k<-12 \Rightarrow k>6.$

   Hence, $6<k<8.$

6. Integral values of $k$

   The only integer in $(6,8)$ is $k=7.$

   Therefore, the number of integral values of $k$ is $1.$

7. Option check

   • A: $2$ ❌
   • B: $0$ ❌
   • C: $1$ ✅
   • D: $3$ ❌

Final Answer: $\boxed{1}$, i.e. Option C.