Solve the following equations by factorization and substitution: 12x2 − 11x + 2 = 0 and 2 imes 22x - 3 imes 2^x + 1 = 0

For $12x^2 - 11x + 2 = 0$, we have $a=12$, $b=-11$, $c=2$.
We need two numbers that multiply to $12 \times 2 = 24$ and add up to $-11$. These numbers are $-3$ and $-8$.

The solutions are $\boxed{x = \frac{1}{4}, \frac{2}{3}}$.

The equation is $2 \cdot (2^x)^2 - 3 \cdot 2^x + 1 = 0$.
Let $y = 2^x$. Substituting this into the equation gives a quadratic equation in terms of $y$.

We need two numbers that multiply to $2 \times 1 = 2$ and add up to $-3$. These numbers are $-1$ and $-2$.

For $y = \frac{1}{2}$:

For $y = 1$:

The solutions are $\boxed{x = -1, 0}$.