For three events, A ,   B and C , P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) =…

$a=\frac{1}{16}$

$c+d+e+f=\frac{1}{4} ...\left(i\right)$

$b+d+e+g=\frac{1}{4} .....\left(ii\right)$

$b+f+c+g=\frac{1}{4} ....\left(iii\right)$

Adding $\left(i\right), \left(ii\right)$ and $\left(iii\right)$

$\Rightarrow 2\left(b+c+d+e+f+g\right)=\frac{3}{4}$

$\Rightarrow b+c+d+e+f+g=\frac{3}{8}$

Now $\Rightarrow P\left(A\cup B\cup C\right)=a+\left(b+c+d+e+f+g\right)$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{1}{16}+\frac{3}{8}$

$\Rightarrow P\left(A\cup B\cup C\right)=\frac{7}{16}$