Find an equation for the line tangent to the curve at the point defined by the given value of $t$. Also, find the value of $\frac{d^{2} y}{d x^{2}}$ at this point.
\[
x=8 \cos t, y=4 \sin t, t=\frac{\pi}{4}
\]

Step 1 ：Given the parametric equations \(x=8 \cos t\) and \(y=4 \sin t\), we are asked to find the equation of the tangent line to the curve at the point defined by \(t=\frac{\pi}{4}\) and the value of the second derivative of \(y\) with respect to \(x\) at this point.

Step 2 ：First, we find the derivative of \(y\) with respect to \(x\), which gives us the slope of the tangent line. We can find this derivative using the chain rule: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Step 3 ：Substituting the given equations into the chain rule, we get \(\frac{dy}{dx} = \frac{4\cos(t)}{-8\sin(t)} = -\frac{\cos(t)}{2\sin(t)}\).

Step 4 ：Next, we find the second derivative of \(y\) with respect to \(x\) by taking the derivative of \(\frac{dy}{dx}\) with respect to \(t\), and then dividing by \(\frac{dx}{dt}\). This gives us \(\frac{d^{2} y}{d x^{2}} = -\frac{1/2 + \cos(t)^{2}/(2\sin(t)^{2})}{8\sin(t)}\).

Step 5 ：Substituting \(t=\frac{\pi}{4}\) into the above equations, we get \(\frac{dy}{dx} = -\frac{1}{2}\) and \(\frac{d^{2} y}{d x^{2}} = -\frac{\sqrt{2}}{8}\).

Step 6 ：Substituting \(t=\frac{\pi}{4}\) into the original parametric equations, we get the point \((x, y) = (4\sqrt{2}, 2\sqrt{2})\).

Step 7 ：Using the point-slope form of a line, \(y - y_{1} = m(x - x_{1})\), where \(m\) is the slope and \((x_{1}, y_{1})\) is a point on the line, we can find the equation of the tangent line to be \(y - 2\sqrt{2} = -\frac{1}{2}(x - 4\sqrt{2})\).

Step 8 ：Final Answer: The equation of the tangent line to the curve at the point defined by \(t=\frac{\pi}{4}\) is \(\boxed{y - 2\sqrt{2} = -\frac{1}{2}(x - 4\sqrt{2})}\). The value of the second derivative of \(y\) with respect to \(x\) at this point is \(\boxed{-\frac{\sqrt{2}}{8}}\).