To solve the expression \((1+\tan^2 25^\circ) \cdot \sin^2 65^\circ\), we will employ trigonometric identities.

First, recall the Pythagorean identity for tangent:
\[
1 + \tan^2 \theta = \sec^2 \theta
\]
So, for \(\theta = 25^\circ\):
\[
1 + \tan^2 25^\circ = \sec^2 25^\circ = \frac{1}{\cos^2 25^\circ}
\]

The expression then becomes:
\[
\frac{1}{\cos^2 25^\circ} \cdot \sin^2 65^\circ
\]

Next, use the co-function identity: \(\sin(90^\circ - \theta) = \cos \theta\). Thus:
\[
\sin 65^\circ = \cos 25^\circ
\]

Substitute \(\sin 65^\circ\) with \(\cos 25^\circ\):
\[
\frac{1}{\cos^2 25^\circ} \cdot (\cos 25^\circ)^2
\]

The \((\cos 25^\circ)^2\) term cancels out the denominator:
\[
\frac{(\cos 25^\circ)^2}{\cos^2 25^\circ} = 1
\]

Thus, the value of the expression is:
\[
\boxed{1}
\]