How to solve recursion $T(n)=T(n/2)+T(n/3)+n$? I do not really know how to approach this kind of recurrence.
This is a straightforward application of the Akra-Bazzi theorem.
You can also show that $T(n) = \Theta(n)$ by induction. First, clearly $T(n) \geq n$. In the other direction, if we have shown that $T(m) \leq 6m$ for $m < n$, then $$ T(n) \leq 6(n/2 + n/3) + n = 6n. $$ Therefore as long as the base case holds, and if we ignore the fact that $n/2,n/3$ are not necessarily integers, we can prove inductively that $T(n) \leq 6n$.