Let's find all numbers $t$ for which exactly one of the following two statements is true:

A) the equation $x^{2}+(t+1) x+1=0$ has two distinct positive roots;

B) the inequality $4 x^{2}+(t-2) x+1 \geq 0$ holds for all positive $x$.

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First, let's examine for which values of $t$ the statement A) holds. The quadratic equation $x^{2}+(t+1) x+1=0$ has two distinct roots precisely when the discriminant of the equation $(t+1)^{2}-4>0$. Furthermore, based on the relationship between the roots and the coefficients, the product of the two roots is 1, meaning that both roots are either positive or both are negative. Thus, the roots of the equation are positive if and only if their sum is positive, that is, if $-(t+1)>0$. Therefore, statement A) holds precisely when $(t+1)^{2}>4$ and $(t+1)<0$, which means $t<-3$.

In the inequality of statement B),

$$
(2 x-1)^{2}+(t+2) x \geq 0
$$

we can see that for $t \geq-2$, on the left side of (1), both terms are non-negative for positive $x$; however, for $t<-2$, the value of the left side of (1) is negative at $x=1 / 2$. Therefore, statement B) holds precisely when $t \geq-2$.

In summary, for $t<-3$, A) is true, and B) is not; for $-3 \leq t<-2$, neither A) nor B) is true; and for $t \geq-2$, B) is true, and A) is not. Thus, the sought values of $t$ are those for which either $t<-3$ or $t \geq-2$.