Let the $A$ be the set of all nonenagative integers.
It is given function such that $f:\mathbb{A}\rightarrow\mathbb{A}$ with $f(1) = 1$ and for every element $n$ od set $A$ following holds:
[b]1)[/b] $3 f(n) \cdot f(2n+1) = f(2n) \cdot (1+3 \cdot f(n))$;
[b]2)[/b] $f(2n) < 6f(n)$, 
Find all solutions of $f(k)+f(l) = 293$, $k<l$.
1. **Analyzing the given functional equations:**

   We start with the given functional equations:
   \[
   3 f(n) \cdot f(2n+1) = f(2n) \cdot (1 + 3 f(n))
   \]
   \[
   f(2n) < 6 f(n)
   \]

2. **Divisibility and bounds:**

   From the first equation, we observe that \(3 f(n)\) must divide \(f(2n)\) because it cannot divide \(1 + 3 f(n)\). Given the second condition \(f(2n) < 6 f(n)\), we deduce:
   \[
   f(2n) = 3 f(n)
   \]
   and
   \[
   f(2n+1) = 3 f(n) + 1
   \]

3. **Modulo properties:**

   From the above, we see that:
   \[
   f(2n) \equiv 0 \pmod{3}
   \]
   \[
   f(2n+1) \equiv 1 \pmod{3}
   \]
   Since \(293 \equiv 2 \pmod{3}\), both \(k\) and \(l\) must be odd.

4. **Base case and induction:**

   We know \(f(1) = 1\). Using the recursive relations:
   \[
   f(2) = 3 f(1) = 3
   \]
   By induction, we can show:
   \[
   f(2^r) = 3^r
   \]
   and
   \[
   f(2^r (2s+1)) = 3^r (3 f(s) + 1)
   \]
   for \(r, s \geq 0\).

5. **Calculating specific values:**

   Using the above relations, we calculate \(f(n)\) for values up to \(f(47)\):
   \[
   f(1) = 1
   \]
   \[
   f(3) = 4
   \]
   \[
   f(5) = 10
   \]
   \[
   f(7) = 13
   \]
   \[
   f(13) = 37
   \]
   \[
   f(15) = 40
   \]
   \[
   f(39) = 256
   \]
   \[
   f(45) = 280
   \]
   \[
   f(47) = 283
   \]

6. **Finding pairs \((k, l)\) such that \(f(k) + f(l) = 293\):**

   We check the pairs:
   \[
   f(5) + f(47) = 10 + 283 = 293
   \]
   \[
   f(7) + f(45) = 13 + 280 = 293
   \]
   \[
   f(13) + f(39) = 37 + 256 = 293
   \]
   \[
   f(15) + f(37) = 40 + 253 = 293
   \]

   These pairs satisfy the condition \(f(k) + f(l) = 293\) with \(k < l\).

The final answer is \( \boxed{ (5, 47), (7, 45), (13, 39), (15, 37) } \)