Let $m$ be equal to $1$ or $2$ and $n<10799$ be a positive integer. Determine all such $n$ for which $\sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}}=m\frac{\sin{n}}{\sin^{2}{1}}$.
1. We start with the given sum:
   \[
   \sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}} = m\frac{\sin{n}}{\sin^{2}{1}}
   \]

2. We can rewrite the sum using the identity for the difference of sines:
   \[
   \frac{1}{\sin{k}\sin{(k+1)}} = \frac{\sin((k+1) - k)}{\sin{k}\sin{(k+1)}} = \frac{\sin(k+1)\cos k - \cos(k+1)\sin k}{\sin{k}\sin{(k+1)}}
   \]

3. This simplifies to:
   \[
   \frac{\sin(k+1)\cos k - \cos(k+1)\sin k}{\sin{k}\sin{(k+1)}} = \cot k - \cot(k+1)
   \]

4. Therefore, the sum becomes a telescoping series:
   \[
   \sum_{k=1}^{n} (\cot k - \cot(k+1))
   \]

5. The telescoping series simplifies to:
   \[
   \cot 1 - \cot(n+1)
   \]

6. Thus, we have:
   \[
   \sum_{k=1}^{n}\frac{1}{\sin{k}\sin{(k+1)}} = \frac{\cot 1 - \cot(n+1)}{\sin 1}
   \]

7. Given that this equals \( m\frac{\sin{n}}{\sin^{2}{1}} \), we equate the two expressions:
   \[
   \frac{\cot 1 - \cot(n+1)}{\sin 1} = m\frac{\sin{n}}{\sin^{2}{1}}
   \]

8. Multiplying both sides by \(\sin^2 1\) gives:
   \[
   (\cot 1 - \cot(n+1)) \sin 1 = m \sin n
   \]

9. Simplifying further:
   \[
   \cot 1 \sin 1 - \cot(n+1) \sin 1 = m \sin n
   \]

10. Since \(\cot 1 \sin 1 = \cos 1\), we have:
    \[
    \cos 1 - \cot(n+1) \sin 1 = m \sin n
    \]

11. Solving for \(\cot(n+1)\):
    \[
    \cot(n+1) = \frac{\cos 1 - m \sin n}{\sin 1}
    \]

12. For \(m = 1\):
    \[
    \cot(n+1) = \frac{\cos 1 - \sin n}{\sin 1}
    \]

13. For \(m = 2\):
    \[
    \cot(n+1) = \frac{\cos 1 - 2 \sin n}{\sin 1}
    \]

14. We need \(\cot(n+1)\) to be a valid cotangent value, which means \(\sin(n+1)\) must be a valid sine value. This implies:
    \[
    \sin(n+1) = 1 \quad \text{or} \quad \sin(n+1) = \frac{1}{2}
    \]

15. Solving \(\sin(n+1) = 1\):
    \[
    n+1 = \frac{\pi}{2} + 2k\pi \quad \text{for integer } k
    \]
    Since \(n\) must be a positive integer less than 10799, we find:
    \[
    n = \frac{\pi}{2} + 2k\pi - 1
    \]

16. Solving \(\sin(n+1) = \frac{1}{2}\):
    \[
    n+1 = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad n+1 = \frac{5\pi}{6} + 2k\pi \quad \text{for integer } k
    \]
    Similarly, we find:
    \[
    n = \frac{\pi}{6} + 2k\pi - 1 \quad \text{or} \quad n = \frac{5\pi}{6} + 2k\pi - 1
    \]

17. We need to check these values to ensure they are less than 10799.

The final answer is \( \boxed{ n = \frac{\pi}{2} + 2k\pi - 1 } \) or \( n = \frac{\pi}{6} + 2k\pi - 1 \) or \( n = \frac{5\pi}{6} + 2k\pi - 1 \) for integer \( k \) such that \( n < 10799 \).