1. Let $d = (2012t+1, 2013t+1)$ then it is easy to see that $d=1$. Thus, both $2012t+1$ and $2013t+1$ are perfect square if and only if $(2012t+1)(2013t+1) = y^2$ for some positive integer $y$. We have
$$\begin{aligned} (2012t+1)(2013t+1) &= y^2 \\ &\Leftrightarrow 4 \cdot 2012^2 \cdot 2013^2 t^2 + 4 \cdot 2012 \cdot 2013 \cdot 4025t + 4 \cdot 2012 \cdot 2013 = 4 \cdot 2012 \cdot 2013 \cdot y^2 \\ &\Leftrightarrow (2 \cdot 2012 \cdot 2013t + 4025)^2 - 1 = 4 \cdot 2012 \cdot 2013 \cdot y^2 \end{aligned}$$
Putting $x = 2 \cdot 2012 \cdot 2013t + 4025$ we have the equation $x^2 - 4 \cdot 2012 \cdot 2013y^2 = 1$.
It is clear that $4 \cdot 2012 \cdot 2013$ is not perfect square, so the Pell equation of 1st type has infinitely many solutions. The fundamental solution of this equation is $(x, y) = (4025, 1)$, thus all its solutions are given by formula
$$\begin{cases} x_0 = 1, x_1 = 4025, x_{n+2} = 8050x_{n+1} - x_n, & n \ge 0. \\ y_0 = 1, y_1 = 1, y_{n+2} = 8050y_{n+1} - y_n & \end{cases}$$
By induction, we can prove that $x_{2i+1}$ is congruent to $4025 \pmod{2 \cdot 2012 \cdot 2013}$ for all $i$ and each value $\frac{x_{2i+1} - 4025}{2 \cdot 2012 \cdot 2013}$ gives us a positive integer $t$ satisfying the required condition.
So there exists infinitely many positive integers $t$ such that $t$ both $2012t+1$ and $2013t+1$ are perfect squares. (Q.E.D)

2. Let $d = (mn+1, mn+n+1)$ then
$$d|(mn+n+1-mn-1) \text{ or } d|n, \text{ therefore } d|(mn+1-mn) \text{ or } d|1.$$
Thus $d=1$ or in other words the numbers $mn+1, (m+1)n+1$ are relatively prime.
So, both $mn+1$ and $(m+1)n+1$ are perfect square if and only if
$$(mn+1)((m+1)n+1) \text{ is perfect square.}$$
Assume that $(mn+1)((m+1)n+1) = y^2$ for $y \in \mathbb{Z}^+$. We have
$$\begin{aligned} & m(m+1)n^2 + (2m+1)n + 1 = y^2 \\ & \Leftrightarrow 4m^2(m+1)^2 n^2 + 4m(m+1)(2m+1)n + 4m(m+1) = 4m(m+1)y^2 \\ & \Leftrightarrow (2m(m+1)n + (2m+1))^2 - 1 = 4m(m+1)y^2 \end{aligned}$$
Putting $x = 2m(m+1)n + (2m+1)$ then we have following equation
$$x^2 - 4m(m+1)y^2 = 1 (*)$$
This is the Pell equation of 1st type. Since $4m(m+1)$ is not perfect square then (*) has infinitely many solutions.
The fundamental solution of equation (*) is $(x, y) = (2m+1, 1)$, so all its solutions $(x_i, y_i)$ can be written in the form
$$\begin{cases} x_0 = 1, x_1 = 2m+1, x_{i+2} = 2(2m+1)x_{i+1} - x_i, & i \ge 0 \\ y_0 = 0, y_1 = 1, y_{i+2} = 2(2m+1)y_{i+1} - y_i \end{cases}$$
By induction, we will prove that $x_{2i}$ is congruent to 1 modulo $2m(m+1)$ and $x_{2i+1}$ is congruent $2m+1$ modulo $2m(m+1)$ for all $i = 0,1,2,...$ (***)
Indeed,
- For $i=0$, by the recurrent formula of $(x_i)$ we see that (*) is true.
- Assume that (*) is true for $i$, i.e $x_{2i}$ is congruent to 1 and $x_{2i+1}$ is congruent $2m+1$ modulo $2m(m+1)$. We have
$$\begin{aligned} x_{2i+2} &= 2(2m+1)x_{2i+1} - x_{2i} \equiv 2(2m+1)(2m+1) - 1 = 8m(m+1) + 1 \equiv 1 \pmod{2m(m+1)}, \\ x_{2i+3} &= 2(2m+1)x_{2i+2} - x_{2i+1} \equiv 2(2m+1) - (2m+1) \equiv 2m+1 \pmod{2m(m+1)}. \end{aligned}$$
Thus, (***) is true for $i+1$.
By induction principle, (***) is true for all $i$.

Next, we will establish the recurrent formula for $r_i = x_{2i+1}$ where $i = 0,1,2,...$.
We have
$$\begin{aligned} r_{i+2} &= x_{2i+5} = 2(2m+1)x_{2i+4} - x_{2i+3} = 2(2m+1)(2(2m+1)x_{2i+3} - x_{2i+2}) - x_{2i+3} \\ &= (4(2m+1)^2 - 1)x_{2i+3} - 2(2m+1)x_{2i+2} = (4(2m+1)^2 - 1)x_{2i+3} - (x_{2i+3} + x_{2i+1}) \\ &= (4(2m+1)^2 - 2)x_{2i+3} - x_{2i+1} = (4(2m+1)^2 - 2)r_{i+1} - r_i \end{aligned}$$
Putting $r_i = 2m(m+1)s_i + (2m+1)$ then the sequence $(s_i), i > 0$ is well-defined and consists of positive integers.
Putting in recurrent formula of $(r_i)$, we get
$$\begin{aligned} 2m(m+1)s_{i+2} + (2m+1) &= (4(2m+1)^2 - 2)(2m(m+1)s_{i+1} + (2m+1)) - (2m(m+1)s_i + (2m+1)) \\ \Leftrightarrow 2m(m+1)s_{i+2} &= 2m(m+1)(4(2m+1)^2 - 2)s_{i+1} - 2m(m+1)s_i + 4(2m+1)(2m+1)^2 - 1 \\ \Leftrightarrow s_{i+2} &= (4(2m+1)^2 - 2)s_{i+1} - s_i + 8(2m+1) \end{aligned}$$
We can compute $r_0 = x_1 = 2m+1$ so $s_0 = 0$ and
$$r_1 = x_3 = 2(2m+1)(2(2m+1)^2 - 1) - (2m+1) = 16m(m+1)(2m+1) + 2m+1,$$
so we have $s_1 = 8(2m+1)$.
We have the recurrent formula for ($s_i$) are
$$\begin{cases} s_0 = 0, s_1 = 8(2m+1), \\ s_{i+2} = (4(2m+1)^2 - 2)s_{i+1} - s_i + 8(2m+1), i \ge 0 \end{cases}$$
From which we can see that all terms of ($s_i$) are divisible by $8(2m+1)$.
Moreover, by putting $x = 2m(m+1)n + (2m+1)$ we can easily see that $n$ satisfies the required condition if and only if $n = s_i, i = 1, 2, 3, ...$ (note that $s_0 = 0$ is not a positive integer).
Thus, all value of $n$ is divisible by $8(2m+1)$. (Q.E.D).