Infinite zeros in infinite series

1. Yes, that is ok.
2. Yes it is. The partial sum of $\sum_k (a_k+b_k)$ is the partial sum of $\sum_ka_k$ plus the partial sum of $\sum_k b_k.$ The result follows from the sum property for limits.
3. Yes. Adding zeros will only delay the inevitable convergence of the sequence of partial sums. Where you insert zeros, the sequence of partial sums will hold flat. For the $N$ you find in the proof of convergence of the original, simply replace with $N$ plus the number of zeros you inserted before the N-th and you'll have the same value that will be within $\epsilon$ of the number the sum converges to.

Both the given identities follow from the fact that $\frac{\pi}{\text{something}}$ is related with the integral over $(0,1)$ of a rational function. The first identity is a consequence of $$\frac{\pi}{4} = \arctan(1)=\int_{0}^{1}\frac{dx}{1+x^2} = \int_{0}^{1}\left(1-x^2+x^4-x^6+\ldots\right)\,dx $$ and for the second series we may perform the same manipulation in the opposite direction, leading to:

$$\begin{eqnarray*}\sum_{n\geq 0}\left(\frac{1}{12n+1}+\frac{1}{12n+5}-\frac{1}{12n+7}-\frac{1}{12n+11}\right)&=&\int_{0}^{1}(1+x^4-x^6-x^{10})\sum_{n\geq 0}x^{12n}\,dx\\&=&\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx\\&=&\int_{0}^{1}\left(\frac{1}{1+x^2}+\frac{x^2}{1+x^6}\right)\,dx\\(x\mapsto z^{1/3})\qquad &=&\frac{\pi}{4}+\frac{1}{3}\int_{0}^{1}\frac{dz}{1+z^2}=\color{red}{\frac{\pi}{3}}.\end{eqnarray*} $$

Indeed, we are just multipling the first series by $\frac{4}{3}$ :D

It is a rewarding exercise to write, or read, detailed rigorous proofs of some simple, "obvious" results.

The definition of $x=\sum_{j=1}^{\infty}y_j$ is that $x=\lim_{n\to \infty}S_n$ where $S_n=\sum_{j=1}^ny_j.$ A useful way to state (or define) that the sequence $S=(S_n)_n$ converges to $x$ is that for any $r>0 $ the set $$F(S, r)=\{n: S_n\not \in [-r+x,r+x]\}$$ is a finite set.

Let $x=\sum_{j=1}^{\infty}y_j.$ Insert some $0$'s into the sequence $(y_j)_j$ to produce a new sequence $(z_i)_i.$ For each $i$ we have either $z_i=0$ or $z_i=y_{g(i)}$ where $g(i)\leq i.$

Let $T=(T_m)_m$ where $T_m=\sum_{i=1}^mz_i.$

Let $i_0$ be the least (or any) $i$ such that $z_i=y_1.$

Then for every $m\geq i_0$ we have $T_m=\sum_{ \{g(i):i\leq m\}}y_{g(i)}= S_{f(m)}$ where $f(m)=\max \{g(i):i\leq m\}.$

Note that for any $n$ there exists $m_0$ such that $z_{m_0}=y_{n+1}$ so there exists $m_0$ such that $f(m_0)=n+1.$

Now for any $r>0$ we have $$F(T,r)=\{m:T_m\not \in [-r+x,r+x]\}\subset \{m\geq i_0:S_{f(m)} \not \in [-r+x.r+x]\}\cup \{m:m<i_0\}$$ $$=\{m\geq i_0: f(m)\in F(S,r)\}\cup \{m:m<i_0\}.$$ Observe that for any $n$ the set $\{m\geq i_0: f(m)=n\}$ is a finite set, because there exists $m_0$ such that $f(m_0)=n+1,$ and hence $m\geq m_0\implies f(m)>n.$ And recall that $F(S,r)$ is a finite set because the sequence $S$ converges to $x.$

So for any $r>0$ the set $F(T,r)$ is a subset of a union of a finite collection of finite sets: $$F(T,r)\subset (\cup_{n\in F(S,r)}\{m\geq i_0:f(m)=n\})\cup \{m:m<i_0\}.$$ Therefore $F(T,r)$ is finite for every $r>0$. Therefore $T$ converges to $x.$ Therefore $x=\sum_{i=1}^{\infty}z_i.$