1A. Solve the inequality

$$
|x+1|^{x^{2}-\frac{5}{2} x+\frac{3}{2}}<1
$$
Solution. First, let's determine the domain. Considering the properties of the exponential function, it is necessary that $|x+1|>0,|x+1| \neq 1$, from which $x \neq -1$ and the domain is $D=\mathbb{R} \backslash\{-1\}$. Given that depending on the base, the exponential function can be both increasing and decreasing, we will consider two cases:

Case 1. Let $0<|x+1|<1$, which means $x \in (-2,0) \backslash \{-1\}$. On this interval, the function is decreasing, so from $|x+1|^{x^{2}-\frac{5}{2} x+\frac{3}{2}}<1 =|x+1|^{0}$, we get $x^{2}-\frac{5}{2} x+\frac{3}{2}>0$. After factoring the quadratic trinomial, we have the inequality $(x-1)\left(x-\frac{3}{2}\right)>0 \Leftrightarrow x \in(-\infty, 1) \cup\left(\frac{3}{2},+\infty\right)$. Finally, after finding the intersection of the intervals, the solution is $x \in(-2,-1) \cup(-1,0)$.

Case 2. Let $|x+1|>1$, that is, $x \in(-\infty,-2) \cup(0,+\infty)$. On this interval, the function is increasing, so from $|x+1|^{x^{2}-\frac{5}{2} x+\frac{3}{2}}<1 =|x+1|^{0}$, it follows that

$$
x^{2}-\frac{5}{2} x+\frac{3}{2}<0 \Leftrightarrow x \in\left(1, \frac{3}{2}\right)
$$

The set of solutions is the union of the solutions obtained in the two cases considered $M=(-2,-1) \cup(-1,0) \cup\left(1, \frac{3}{2}\right)$.