For any natural numbers $$m, n$$, and $$k$$, such that $$k$$ divides both $$m + 2n$$ and $$3m + 4n$$, $$k$$ must be a common divisor of:

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1. **Given conditions**:  
   - $$ k $$ divides $$ m + 2n $$.  
   - $$ k $$ divides $$ 3m + 4n $$.

2. **Finding relationships**:  
   Since $$ k $$ divides both expressions, it must also divide their linear combinations. Consider:  
   - Subtracting twice the first equation from the second:  
     $$
     (3m + 4n) - 2(m + 2n) = m.
     $$
     Hence, $$ k $$ divides $$ m $$.

- Subtracting the first equation from the second:  
     $$
     (3m + 4n) - (m + 2n) = 2m + 2n.
     $$
     Dividing by 2 gives:  
     $$
     m + n.
     $$

Combining these results, since $$ k $$ divides both $$ m + 2n $$, it must also divide:  
   - The difference:  
     $$
     (m + 2n) - m = 2n.
     $$

Thus, $$ k $$ divides both $$ m $$ and $$ 2n $$. Therefore, the correct answer is **$$ m $$ and $$ 2n $$**.

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