Since \( a \) varies inversely as \( b^2 \), we can express this relationship as:
\[ a = \frac{k}{b^2} \]
where \( k \) is a constant.
We know that \( a = 25 \) when \( b = \frac{4}{5} \). Using this information, we can solve for \( k \):
\[ 25 = \frac{k}{\left(\frac{4}{5}\right)^2} \]
Calculating \( \left(\frac{4}{5}\right)^2 \):
\[ \left(\frac{4}{5}\right)^2 = \frac{16}{25} \]
Now, substituting this back into the equation gives:
\[ 25 = \frac{k}{\frac{16}{25}} \]
To isolate \( k \), we multiply both sides by \( \frac{16}{25} \):
\[ k = 25 \cdot \frac{16}{25} = 16 \]
Now that we have the value of \( k \), we can use it to find \( a \) when \( b = \frac{7}{2} \):
\[ a = \frac{16}{\left(\frac{7}{2}\right)^2} \]
Calculating \( \left(\frac{7}{2}\right)^2 \):
\[ \left(\frac{7}{2}\right)^2 = \frac{49}{4} \]
Now substituting this back into the equation for \( a \):
\[ a = \frac{16}{\frac{49}{4}} = 16 \cdot \frac{4}{49} = \frac{64}{49} \]
Thus, the value of \( a \) when \( b = \frac{7}{2} \) is:
\[ \boxed{\frac{64}{49}} \]