4. Given a geometric sequence $\left\{b_{n}\right\}$ with all terms being positive and an arithmetic sequence $\left\{a_{n}\right\}$, satisfying $b_{3}-b_{1}=9, b_{5}-b_{3}=36$, and $b_{1}=$ $a_{1}, b_{2}=a_{3}$. Let the sum of the first 6 terms of the geometric sequence be $G_{6}$, and the sum of the first 12 terms of the arithmetic sequence be $A_{12}$, then $G_{6}+A_{12}=$ $\qquad$ .
4. 324.

Let the common ratio be $q(q>0)$, then
$$
\left\{\begin{array}{l}
b_{1} q^{2}-b_{1}=9, \\
b_{1} q^{4}-b_{1} q^{2}=36,
\end{array}\right.
$$

which is equivalent to $\left\{\begin{array}{l}b_{1} q^{2}-b_{1}=9, \\ q^{2}\left(b_{1} q^{2}-b_{1}\right)=36 .\end{array}\right.$
Substituting (1) into (2) gives $q^{2}=4$. Hence, $q=2$ or $q=-2$. Since $q>0$, then $q=2$. At this point, $b_{1}=3$.
Therefore, $G_{6}=\frac{b_{1}\left(1-q^{6}\right)}{1-q}=\frac{3(1-64)}{1-2}=189$.
Let $d$ be the common difference of the arithmetic sequence $\left\{a_{n}\right\}$, then according to the given conditions, $\left\{\begin{array}{l}a_{1}=3, \\ a_{1}+2 d=6 .\end{array}\right.$ This yields $d=\frac{3}{2}$. Therefore,
$$
a_{12}=a_{1}+11 d=\frac{39}{2} \text {. }
$$

Thus, $A_{12}=\frac{\left(a_{1}+a_{12}\right) \times 12}{2}=\left(3+\frac{39}{2}\right) \times 6=135$.
Hence, $G_{6}+A_{12}=189+135=324$.