Part (a)
$w_o = 62.5(9s) = 562.5s \, \text{lb/ft}$

$F_w = \frac{1}{2} w_o(9)$

$F_w = \frac{1}{2} (562.5s)(9)$

$F_w = 2531.25s \, \text{lb}$

$\Sigma M_{R1} = 0$

$15R_2 = 12F_w$

$15R_2 = 12(2531.25s)$

$R_2 = 2025s$

$\Sigma M_{R2} = 0$

$15R_1 = 3F_w$

$15R_1 = 3(2531.25s)$

$R_1 = 506.25s$

Location of Maximum Moment

$\dfrac{y}{x} = \dfrac{562.5s}{9}$

$y = 62.5s$

$506.25s - \frac{1}{2}xy = 0$

$506.25s - \frac{1}{2}x (62.5sx) = 0$

$x^2 = 16.2$

$x = 4.02 \, \text{ft}$

Maximum Moment

$M = (506.25s)(6) + \frac{2}{3}(x)(506.25s)$

$M = 3037.5s + 337.5(4.02s)$

$M = 4394.25s$

Required Spacing

$(\,f_b\,)_{max} = \dfrac{Mc}{I}$

$100 = \dfrac{4394.25s(12)(12/2)}{\frac{8(12^3)}{12}}$

$s = 3.64 \, \text{ft}$ *answer*

Part (b)

$w_o = 62.5(15)(3.64) = 3412.5 \, \text{lb/ft}$

$F_w = \frac{1}{2}w_o(15)$

$F_w = \frac{1}{2}(3412.5)(15)$

$F_w = 25\,593.75 \, \text{lb}$

$\Sigma M_{R1} = 0$

$15R_2 = 10F_w$

15R_2 = 10(25\,593.75)

$R_2 = 17\,062.5 \, \text{lb}$

$\Sigma M_{R2} = 0$

$15R_1 = 5F_w$

$15R_1 = 5(25\,593.75)$

$R_1 = 8\,531.25 \, \text{lb}$

Location of Maximum Moment (Shear = 0)

$\dfrac{y}{x} = \dfrac{3412.5}{15}$

$y = 227.5x$

$8531.25 - \frac{1}{2}xy = 0$

$8531.25 - \frac{1}{2}x(227.5x) = 0$

$x^2 = 75$

$x = 8.66 \, \text{ft}$

Maximum Moment

$M = \frac{2}{3}x(8531.25)$

$M = \frac{2}{3}(8.66)(8531.25)$

$M = 49\,255.19 \, \text{lb}\cdot\text{ft}$

Actual Stress

$f_b = \dfrac{Mc}{I}$

$f_b = \dfrac{(49\,255.19)(12)(12/2)}{\dfrac{8(12^3)}{12}}$

$f_b = 3\,078.36 \, \text{psi} \gt 1600 \, \text{psi}$

Therefore, the 3.64 ft spacing of timbers is *not safe* when water reaches its maximum depth of 15 ft.