Solution:
$\frac{45}{11}$

Using long division, when $a x^{3}+b x^{2}+c x+5$ is divided by $x^{2}+x+2$, the quotient is $a x+(b-a)$ and the remainder is $(c-a-b) x+5+2 a-2 b$. Since $x^{2}+x+2$ is a factor of $a x^{3}+b x^{2}+c x+5$, we must have $c-a-b=0$ and $5+2 a-2 b=0$. On the other hand, since $2 x-1$ is a factor of $a x^{3}+b x^{2}+c x-\frac{25}{16}$, by the Remainder Theorem, we must have
$$a\left(\frac{1}{2}\right)^{3}+b\left(\frac{1}{2}\right)^{2}+c\left(\frac{1}{2}\right)-\frac{25}{16}=0$$
or
$$\frac{a}{8}+\frac{b}{4}+\frac{c}{2}-\frac{25}{16}=0 \quad \text{ or } \quad 2 a+4 b+8 c-25=0$$
Solving the following system of equations:
$$\left\{\begin{aligned} c-a-b & =0 \\ 5+2 a-2 b & =0 \\ 2 a+4 b+8 c-25 & =0 \end{aligned}\right.$$
we get $a=-\frac{5}{22}$, $b=\frac{25}{11}$, and $c=\frac{45}{22}$, so that $a+b+c=\frac{45}{11}$.