Geometry

The answer is 2/9.

Law of Cosines

AC^2  = BA^2 + BC^2  - 2 (BA * BC) cos B

7^2  = 5^2 + 32  - 2 (5 * sqrt 32) cos B

(-8) / (-10 sqrt 32)  = cos B

(-8) / (- 5 * 2 * 4 sqrt 2)  = cos B

1 / (5 sqrt 2)  = cos B

We are given:
[
\sin A = \frac{4}{5}, \quad \tan C = 1
]

Step 1: Find (\cos A)

Since (\sin A = \frac{4}{5}), we use:
[
\cos A = \frac{3}{5}
]
(assuming (A) is acute, as is standard in triangle problems)

Step 2: Find (\sin C) and (\cos C)

[
\tan C = 1 \Rightarrow C = 45^\circ
]
So:
[
\sin C = \cos C = \frac{\sqrt{2}}{2}
]

Step 3: Use angle sum

[
B = 180^\circ - (A + C)
]
So:
[
\cos B = -\cos(A + C)
]

Step 4: Compute (\cos(A + C))

[
\cos(A + C) = \cos A \cos C - \sin A \sin C
]
[
= \frac{3}{5}\cdot \frac{\sqrt{2}}{2} - \frac{4}{5}\cdot \frac{\sqrt{2}}{2}
= \frac{\sqrt{2}}{10}(3 - 4)
= -\frac{\sqrt{2}}{10}
]

Step 5: Final answer

[
\cos B = -(-\frac{\sqrt{2}}{10}) = \frac{\sqrt{2}}{10}
]

✅ Final Answer:
[
\boxed{\frac{\sqrt{2}}{10}}