(b) $$ (2 x-3 y)^{5} $$ $$ r=2, a=2 x $$

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To find the term corresponding to $$ r=2 $$ in the expansion of 
$$
(2x - 3y)^5,
$$
with $$ a=2x $$, we use the binomial theorem. The general term in the expansion of 
$$
(a + b)^n
$$
is given by
$$
T_{r+1} = \binom{n}{r} \, a^{\,n-r} \, b^r.
$$

In this problem, we have:
- $$ n = 5 $$,
- $$ a = 2x $$,
- $$ b = -3y $$,
- $$ r = 2 $$.

Thus, the term when $$ r = 2 $$ (i.e. the third term in the expansion) is:
$$
T_{3} = \binom{5}{2} (2x)^{5-2} (-3y)^{2}.
$$

Now, we calculate each part step by step:

1. Calculate the binomial coefficient:
   $$
   \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10.
   $$

2. Compute $$(2x)^{5-2} = (2x)^{3}$$:
   $$
   (2x)^3 = 2^3 \, x^3 = 8x^3.
   $$

3. Compute $$(-3y)^{2}$$:
   $$
   (-3y)^2 = (-3)^2 \, y^2 = 9y^2.
   $$

4. Combine the results to obtain the term:
   $$
   T_{3} = 10 \times 8x^3 \times 9y^2.
   $$
   Multiplying the numerical coefficients:
   $$
   10 \times 8 \times 9 = 720.
   $$

Thus, the term corresponding to $$ r = 2 $$ is:
$$
T_{3} = 720x^3y^2.
$$
The term corresponding to $$ r = 2 $$ in the expansion of $$ (2x - 3y)^5 $$ is $$ 720x^3y^2 $$.

(b)

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