Using three-moment equation

$M_1 L_1 + 2M_2(L_1 + L_2) + M_3 L_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 0$

Where,

$M_1 = M_3 = 0$
$L_1 = L_2 = L$

$\dfrac{6A_1\bar{a}_1}{L_1} = \dfrac{6A_2\bar{b}_2}{L_2} = \frac{1}{4}w_o L^3$

Thus,

$0 + 2M_2(L + L) + 0 + \frac{1}{4}w_o L^3 + \frac{1}{4}w_o L^3 = 0$

$4L \, M_2 + \frac{1}{2}w_o L^3$

$M_2 = -\frac{1}{8}w_o L^2$ *answer*

**From the first span**

Simple reactions due to loadings

$V_{1L} = V_{1R} = \frac{1}{2}w_o L$

Couple reaction due to end moment

${R_1}' = \dfrac{M_2}{L} = \dfrac{\frac{1}{8}w_o L^2}{L}$

${R_1}' = \frac{1}{8}w_o L$

Thus,

$R_{1L} = V_{1L} - {R_1}' = \frac{1}{2}w_o L - \frac{1}{8}w_o L$

$R_{1L} = \frac{3}{8}w_o L$

$R_{1R} = V_{1R} + {R_1}' = \frac{1}{2}w_o L + \frac{1}{8}w_o L$

$R_{1R} = \frac{5}{8}w_o L$

**From the second span**

Simple reactions due to loadings

$V_{2L} = V_{2R} = \frac{1}{2}w_o L$

Couple reaction due to end moment

${R_2}' = \dfrac{M_2}{L} = \dfrac{\frac{1}{8}w_o L^2}{L}$

${R_2}' = \frac{1}{8}w_o L$

Thus,

$R_{2L} = V_{2L} + {R_2}' = \frac{1}{2}w_o L + \frac{1}{8}w_o L$

$R_{2L} = \frac{5}{8}w_o L$

$R_{2R} = V_{2R} - {R_2}' = \frac{1}{2}w_o L - \frac{1}{8}w_o L$

$R_{2R} = \frac{3}{8}w_o L$

Note: You can actually use the 'symmetry' to solve for R_{2L} and R_{2R}. It is easy easy to see that R_{2L} = R_{1R} and R_{2R} = R_{1L}. With this, you can shorten the solution by not doing all computations related to the second span.

**From the load diagram**

$R_1 = \frac{3}{8}w_o L$ *answer*

$R_2 = \frac{5}{8}w_o L + \frac{5}{8}w_o L = \frac{5}{4}w_o L$ *answer*

$R_3 = \frac{3}{8}w_o L$ *answer*

**From the shear diagram**

By ratio and proportion

$\dfrac{x}{\frac{3}{8}w_o L} = \dfrac{L - x}{\frac{5}{8}w_o L}$

$5x = 3L - 3x$

$x = \frac{3}{8}L$

$M_{max\,(+)} = \frac{1}{2}x (\frac{3}{8}w_o L) = \frac{1}{2}( \frac{3}{8}L)(\frac{3}{8}w_o L)$

$M_{max\,(+)} = \frac{9}{128}w_o L^2$ *answer*