How to find a good asymptotic approximation of T(n) = T(n/2) + T(n/3) + 1?
You can write your recurrence as
$$
T(n) = \sum_{i=1}^k a_i T(b_i x + h_i(n))  + g(n)
$$
with:


$k=2$
$a_1 = a_2 = 1$
$b_1 = \frac{1}{2}$, and $b_2 = \frac{1}{3}$
$h_1(n) = h_2(n) = 0$
$g(n) = 1$


From Akra–Bazzi theorem, the solution to your recurrence is $T(n) = \Theta\Big( n^p \big(1 + f(n)\big)\Big)$, where $p$ is such that $a_1 b_1^p + a_2 b_2^p =1$ and $f(n) = \int_1^n \frac{1}{u^{p+1}} \text{d} u$. 

Substituting, we have $2^{-p} + 3^{-p} = 1$, which shows that $0.78 &lt; p &lt; 0.79$.

Therefore: $$
f(n) = \int_1^n \frac{1}{u^{p+1}} \text{d} u \le \int_1^\infty \frac{1}{u^{p+1}} \text{d} u = \frac{1}{p} = \Theta(1),
$$

and the solution to your recurrence is $T(n) = \Theta( n^p )$.