Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
1. $f(1) = 1$;
2. For all $n \in \mathbb{N}$, $3f(n) f(2n+1) = f(2n) (1 + 3f(n))$;
3. For all $n \in \mathbb{N}$, $f(2n) < 6 f(n)$.

We need to find all solutions of the equation $f(k) + f(l) = 293$ where $k < l$.

By induction, we can show that $f(2n) = 3f(n)$ and $f(2n+1) = 3f(n) + 1$. Using these properties, we can compute the values of $f(n)$ for various $n$:

$$\begin{aligned} f(1) &= 1, \\ f(2) &= 3, \\ f(3) &= 4, \\ f(4) &= 9, \\ f(5) &= 10, \\ f(6) &= 12, \\ f(7) &= 13, \\ f(8) &= 27, \\ f(9) &= 28, \\ f(10) &= 30, \\ f(11) &= 31, \\ f(12) &= 36, \\ f(13) &= 37, \\ f(14) &= 39, \\ f(15) &= 40, \\ f(16) &= 81, \\ f(17) &= 82, \\ f(18) &= 84, \\ f(19) &= 85, \\ f(20) &= 90, \\ f(21) &= 91, \\ f(22) &= 93, \\ f(23) &= 94, \\ f(24) &= 108, \\ f(25) &= 109, \\ f(26) &= 111, \\ f(27) &= 112, \\ f(28) &= 117, \\ f(29) &= 118, \\ f(30) &= 120, \\ f(31) &= 121, \\ f(32) &= 243, \\ f(33) &= 244, \\ f(34) &= 246, \\ f(35) &= 247, \\ f(36) &= 252, \\ f(38) &= 255, \\ f(39) &= 256, \\ f(40) &= 270, \\ f(41) &= 271, \\ f(42) &= 273, \\ f(43) &= 274, \\ f(44) &= 279, \\ f(45) &= 280, \\ f(46) &= 282, \\ f(47) &= 283. \end{aligned}$$

From these values, we find the pairs $(k, l)$ such that $f(k) + f(l) = 293$ and $k < l$. The valid pairs are:
$$(5, 47), (7, 45), (13, 39), (15, 37).$$

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