It is given that the first three terms, in ascending powers of $x$, of the binomial expansion of $(2+ax)^6$ are $64 - 960x + bx^2$ respectively.

1. Expand the first three terms of $(2+ax)^6$ using the binomial theorem

The first three terms of the expansion of $(2+ax)^6$ are:
$2^6 + \binom{6}{1} 2^5 (ax) + \binom{6}{2} 2^4 (ax)^2$
$= 64 + 6(32)(ax) + 15(16)(a^2 x^2)$
$= 64 + 192ax + 240a^2 x^2$

2. Compare the coefficient of $x$ to find $a$

Comparing the coefficient of $x$ with the given expansion $64 - 960x + bx^2$:
$192a = -960$
$a = -5$

3. Compare the coefficient of $x^2$ to find $b$

Comparing the coefficient of $x^2$:
$b = 240a^2$
Substitute $a = -5$:
$b = 240(-5)^2$
$b = 240(25) = 6000$

1. Retrieve the value of $a$ from part (a) and find the $x^3$ term in the expansion of $(2+ax)^6$.

From part (a), $a = -5$.
The term in $x^3$ for $(2-5x)^6$ is:
$\binom{6}{3}(2)^3(-5x)^3 = 20(8)(-125x^3) = -20000x^3$

2. Expand $(1+3x^2)^5$ up to the term in $x^2$.

$(1+3x^2)^5 = 1^5 + \binom{5}{1}(1)^4(3x^2) + \dots = 1 + 15x^2 + \dots$

3. Identify and multiply the terms that produce $x^3$ in the product.

We need the $x^3$ terms from the product:
$(1 + 15x^2 + \dots)(64 - 960x + 6000x^2 - 20000x^3 + \dots)$

The terms in $x^3$ are:
$(1)(-20000x^3) + (15x^2)(-960x)$
$= -20000x^3 - 14400x^3 = -34400x^3$