1. Determine all values of the real parameter $a$ for which the equation

$$
a 2^{a x^{2}}-2^{a+x^{2}}=32
$$

has exactly two solutions in the set of real numbers, and that these solutions differ by 2.
1. Note that if $x$ is a solution to the equation in question (for some fixed parameter $a$), then $-x$ is also a solution. Therefore, if the parameter $a$ satisfies the conditions in the problem, the real solutions of the given equation must be the numbers 1 and -1. Now, using the fact that $x=1$ is a solution, we get $a 2^{a}-2^{a+1}=32$, i.e., $2^{a}(a-2)=32$. Since $2^{a}$ is positive, it follows that $a-2$ must also be positive, i.e., $a>2$. However, in this case, both $2^{a}$ and $a-2$ are positive and increasing functions, so the left side of the observed expression is an increasing function, and there can be at most one such parameter $a$. Clearly, $a=4$ satisfies the observed equality, so $a=4$ is the only possible value of the parameter $a$, and it remains to verify whether the conditions in the problem are indeed satisfied in this case.

The given equation then reduces to $4 \cdot 2^{4 x^{2}}-2^{4+x^{2}}=32$, i.e., $2^{4 x^{2}}-2^{2+x^{2}}=8$. After substituting $y=x^{2}$, we get $2^{4 y}=8+2^{2+y}$, which is equivalent to $2^{3 y}=\frac{8}{2^{y}}+4$. Note that for non-negative $y$ (and $y=x^{2} \geqslant 0$), the left side is an increasing function of $y$, while the right side is a decreasing function. Therefore, there is at most one $y$ that satisfies this equality, and since $y=1$ clearly satisfies it, this is the only solution to the observed equation. Thus, in the original equation, $x^{2}=1$, i.e., $x= \pm 1$, which means that the original equation indeed has exactly two real solutions and that they differ by 2.

Therefore, the only value of the parameter $a$ that satisfies the conditions of the problem is $a=4$.