Given A = k(B2/C2), solve for A and percentage change in A under specified conditions.

Substitute these values into the equation:

$$20 = k \frac{5^2}{9^2}$$ $$20 = k \frac{25}{81}$$

To solve for $k$, multiply both sides by $\frac{81}{25}$:

$$k = 20 \times \frac{81}{25}$$ $$k = \frac{4 \times 5 \times 81}{5 \times 5}$$ $$k = \frac{4 \times 81}{5}$$ $$k = \frac{324}{5}$$ $$k = 64.8$$

Substitute $k=64.8$, $B=7$, and $C=25$ into the equation:

$$A = 64.8 \frac{7^2}{25^2}$$ $$A = 64.8 \frac{49}{625}$$ $$A = \frac{64.8 \times 49}{625}$$ $$A = \frac{3175.2}{625}$$ $$A = 5.08032$$

Rounding to a reasonable number of decimal places, for example, two:

$$A \approx 5.08$$

The value of A when B=7 and C=25 is $\boxed{5.08}$.

Let the original values be $B_1$ and $C_1$.
If B is increased by 20%, the new value $B_2$ is:

$$B_2 = B_1 + 0.20 B_1 = 1.20 B_1$$

If C is decreased by 36%, the new value $C_2$ is:

$$C_2 = C_1 - 0.36 C_1 = 0.64 C_1$$

Original A ($A_1$):

$$A_1 = k \frac{B_1^2}{C_1^2}$$

New A ($A_2$):

$$A_2 = k \frac{B_2^2}{C_2^2}$$

Substitute $B_2 = 1.20 B_1$ and $C_2 = 0.64 C_1$ into the equation for $A_2$:

$$A_2 = k \frac{(1.20 B_1)^2}{(0.64 C_1)^2}$$ $$A_2 = k \frac{1.44 B_1^2}{0.4096 C_1^2}$$ $$A_2 = \frac{1.44}{0.4096} \left( k \frac{B_1^2}{C_1^2} \right)$$

Recognize that $k \frac{B_1^2}{C_1^2}$ is $A_1$:

$$A_2 = \frac{1.44}{0.4096} A_1$$ $$A_2 = 3.515625 A_1$$

Percentage change = $\frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$

$$\text{Percentage change in A} = \frac{A_2 - A_1}{A_1} \times 100\%$$ $$\text{Percentage change in A} = \frac{3.515625 A_1 - A_1}{A_1} \times 100\%$$ $$\text{Percentage change in A} = (3.515625 - 1) \times 100\%$$ $$\text{Percentage change in A} = 2.515625 \times 100\%$$ $$\text{Percentage change in A} = 251.5625\%$$

Rounding to two decimal places:

$$\text{Percentage change in A} \approx 251.56\%$$

The percentage change in A is an increase of $\boxed{251.56\%}$.