In the binomial expansion \( (x + 1)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} \), the terms are: \( \binom{7}{0} x^7 = x^7 \), \( \binom{7}{1} x^6 = 7x^6 \), \( \binom{7}{2} x^5 = bx^5 \), \( \binom{7}{3} x^4 = 35x^4 \), …, \( \binom{7}{7} = 1 \).

For the \( x^5 \)-term (\( k = 2 \)): \( \binom{7}{2} = \frac{7 \cdot 6}{2 \cdot 1} = 21 \). Thus, \( b = 21 \).

Verify: \( \binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35 \), matches the given \( 35x^4 \).

Answer: \( b = 21 \)

Terms: 2nd: \( 7x^6 \), 3rd: \( 21x^5 \), 4th: \( 35x^4 \). Given: \( 21x^5 = \frac{7x^6 + 35x^4}{2} \).

Multiply by 2: \( 42x^5 = 7x^6 + 35x^4 \).

Rearrange: \( 7x^6 + 35x^4 – 42x^5 = 0 \).

Factor: \( x^4 (7x^2 – 42x + 35) = 0 \). Since \( x \neq 0 \), solve: \( 7x^2 – 42x + 35 = 0 \).

Discriminant: \( \Delta = (-42)^2 – 4 \cdot 7 \cdot 35 = 1764 – 980 = 784 \).

\[ x = \frac{42 \pm \sqrt{784}}{14} = \frac{42 \pm 28}{14} \]

\[ x = \frac{70}{14} = 5, \quad x = \frac{14}{14} = 1 \]

Verify: For \( x = 1 \): \( 21 \cdot 1^5 = 21 \), \( \frac{7 \cdot 1^6 + 35 \cdot 1^4}{2} = \frac{7 + 35}{2} = 21 \). For \( x = 5 \): \( 21 \cdot 5^5 = 65625 \), \( \frac{7 \cdot 5^6 + 35 \cdot 5^4}{2} = \frac{7 \cdot 15625 + 35 \cdot 625}{2} = 65625 \).

Answer: \( x = 1, 5 \)