If the foci of the ellipse `(x^(2))/(16)+(y^(2))/(b^(2))=1` and the hyperbola `(x^(2))/(144)-(y^(2))/(81)=(1)/(125)` coincide, the find the value of `b^(2)`.

Explanation

To solve the problem, we need to find the value of \( b^2 \) such that the foci of the given ellipse and hyperbola coincide.

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse is given by:
\[ \frac{x^2}{16} + \frac{y^2}{b^2} = 1 \]
Here, \( a^2 = 16 \) and \( b^2 = b^2 \).

### Step 2: Calculate the distance of the foci of the ellipse
The distance of the foci from the center for an ellipse is given by:
\[ c_e = \sqrt{a^2 - b^2} \]
Substituting \( a^2 = 16 \):
\[ c_e = \sqrt{16 - b^2} \]

### Step 3: Identify the parameters of the hyperbola
The equation of the hyperbola is given by:
\[ \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{125} \]
To rewrite it in standard form, we multiply through by 125:
\[ \frac{125x^2}{144} - \frac{125y^2}{81} = 1 \]
This gives us:
\[ \frac{x^2}{\frac{144}{125}} - \frac{y^2}{\frac{81}{125}} = 1 \]
From this, we can identify \( a^2 = \frac{144}{125} \) and \( b^2 = \frac{81}{125} \).

### Step 4: Calculate the distance of the foci of the hyperbola
The distance of the foci from the center for a hyperbola is given by:
\[ c_h = \sqrt{a^2 + b^2} \]
Substituting \( a^2 = \frac{144}{125} \) and \( b^2 = \frac{81}{125} \):
\[ c_h = \sqrt{\frac{144}{125} + \frac{81}{125}} = \sqrt{\frac{144 + 81}{125}} = \sqrt{\frac{225}{125}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} \]

### Step 5: Set the distances equal
Since the foci of the ellipse and hyperbola coincide, we set \( c_e = c_h \):
\[ \sqrt{16 - b^2} = \frac{3}{\sqrt{5}} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 16 - b^2 = \frac{9}{5} \]

### Step 7: Solve for \( b^2 \)
Rearranging the equation:
\[ b^2 = 16 - \frac{9}{5} \]
To combine the terms, convert 16 to a fraction:
\[ 16 = \frac{80}{5} \]
Thus,
\[ b^2 = \frac{80}{5} - \frac{9}{5} = \frac{71}{5} \]

### Final Answer
The value of \( b^2 \) is:
\[ \boxed{\frac{71}{5}} \]