Take care of just 1 and 2

What we want to find out is:-

$\displaystyle \sum_{r=0}^{\infty}\frac{1}{(3r+1)(3r+2)} + 2\sum_{r=0}^{\infty}\frac{1}{(3r+2)(3r+3)}$

both of these series are convergent by Limit comparison Test.

We know $\displaystyle \frac{1}{1-y} = \sum_{r=0}^{\infty}y^{r}$ , $\displaystyle 0\leq y <1$

So $\displaystyle \frac{1}{1-y^{3}} = \sum_{r=0}^{\infty}y^{3r}$

So $\displaystyle \int_{0}^{x}\frac{dy}{1-y^{3}} = \sum_{r=0}^{\infty}\frac{x^{3r+1}}{3r+1}$ , $\displaystyle 0\leq x < 1$

So $\displaystyle \int_{0}^{1}\int_{0}^{x}\frac{1}{1-y^{3}}dydx = \sum_{r=0}^{\infty}\frac{1}{(3r+1)(3r+2)}$

Again :- $\displaystyle \frac{y}{1-y^{3}} = \sum_{r=0}^{\infty}y^{3r+1}$

So $\displaystyle \int_{0}^{x}\frac{ydy}{1-y^{3}} = \sum_{r=0}^{\infty}\frac{x^{3r+2}}{3r+2}$

So $\displaystyle \int_{0}^{1}\int_{0}^{x}\frac{y}{1-y^{3}}dydx = \sum_{r=0}^{\infty}\frac{1}{(3r+2)(3r+3)}$

So we just have to evaluate the integral :-

$\large \int_{0}^{1}\int_{0}^{x} \frac{1+2y}{1-y^{3}} dydx$

Now what we want to do is change the order of integration:-

So the integral becomes

$\large \int_{0}^{1}\int_{y}^{1} \frac{1+2y}{1-y^{3}} dxdy$

$\large = \int_{0}^{1} \frac{(1+2y)(1-y)}{1-y^{3}}dy$

$\large = \int_{0}^{1} \frac{1+2y}{1+y+y^{2}}dy$

Now substituting $1+y+y^{2} = z$ we have $(1+2y)dy = dz$

So our integral becomes:-

$\large \int_{1}^{3} \frac{dz}{z} = \ln(3)$ which is our answer.

Consider the Maclaurin series below:

$\begin{aligned} \frac 1{1-x} & = \sum_{n=0}^\infty x^n & \small \blue{\text{Integrate w.r.t. }x} \\ - \ln (1-x) & = \sum_{n=1}^\infty \frac {x^n}{n} & \small \blue{\text{Integrate w.r.t. }x \text{ again}} \\ (1-x)\ln(1-x)+x & = \sum_{n=1}^\infty \frac {x^{n+1}}{n(n+1)} \end{aligned}$

Note that in both cases, the constant of integration $C = 0$ , because $LHS = 0$ , when $x=0$ .

Now let $\displaystyle S(x) = \sum_{n=1}^\infty \frac {x^{n+1}}{n(n+1)} = (1-x)\ln(1-x)+x$ . We find that, where $\omega$ is the cubic root of unity:

$\begin{aligned} S(1) + S(\omega) + S(\omega^2) & = \frac 3{2(3)} + \frac 3{5(6)} + \frac 3{8(9)} + \cdots \\ S(1) + \omega S(\omega) + \omega^2 S(\omega^2) & = \frac 3{1(2)} + \frac 3{4(5)} + \frac 3{7(8)} + \cdots \end{aligned}$

Therefore,

$\begin{aligned} \sum_{n=1}^\infty \frac {n \bmod 3}{n(n+1)} & = \frac {3S(1) + (\omega + 2)S(\omega) + (\omega^2 + 2)S(\omega^2)}3 \\ & = \small \frac 13 \left(3\lim_{x \to 1}(1-x)\ln(1-x) + 3 + (\omega + 2)\left((1-\omega) \ln (1-\omega) + \omega \right) + (\omega^2 + 2)\left((1-\omega^2) \ln (1-\omega^2) + \omega^2 \right) \right) \\ & = \frac 13 \left((\omega + 2)(1-\omega) \ln (1-\omega) + (\omega^2 + 2)(1-\omega^2) \ln (1-\omega^2) \right) \\ & = \frac 13 \left((1-\omega^2)(1-\omega) \ln (1-\omega) + (1-\omega)(1-\omega^2) \ln (1-\omega^2) \right) \\ & = \frac 13 \left((1-\omega)(1-\omega^2) \ln ((1-\omega)(1-\omega^2))\right) \\ & = \frac 13 \left(3 \ln 3\right) \\ & = \ln 3 \end{aligned}$

And $a^a = 3^3 = \boxed {27}$ .