We know that the product of the real numbers $a$ and $b$ is 1, and

$$
\frac{a+b+2}{4}=\frac{1}{a+1}+\frac{1}{b+1}
$$

Determine the values of $a$ and $b$.
I. solution. Due to the denominators of the fractions, $a \neq-1$ and $b \neq-1$. Rearrange the equation:

$$
\frac{a+b+2}{4}=\frac{b+1+a+1}{(a+1)(b+1)}, \quad(a+b+2)(a+1)(b+1)=4(a+b+2)
$$

$(a+1)(b+1)=a b+a+b+1=a+b+2$, since $a b=1$. Thus

$$
(a+b+2)(a+b+2)=4(a+b+2), \quad \text { hence } \quad(a+b+2)(a+b-2)=0
$$

The first factor, as we have seen, is now $(a+1)(b+1)$, which cannot be 0 due to the condition.

If $a+b-2=0$, then based on $b=\frac{1}{a}$, we have $a+\frac{1}{a}-2=0$, from which $(a-1)^{2}=0$, so $a=1, b=1$. It is easily verifiable that this is indeed a solution to the problem.

II. solution. Multiply both sides of equation (1) by $4(a+1)(b+1)$ $(a \neq-1, b \neq-1):(a+b+2)(a+$ $1)(b+1)=4(a+b+2)$. If we perform the operations and replace $a b$ with 1 everywhere, we arrive at the equation $a^{2}+b^{2}=2$. Solve the following system of equations:

$$
\left.\begin{array}{rl}
a^{2}+b^{2} & =2 \\
a b & =1
\end{array}\right\}
$$

First, add twice the second equation to the first equation, then subtract the first equation from the second equation. Then we get:

$$
\left.\begin{array}{l}
(a+b)^{2}=4 \\
(a-b)^{2}=0
\end{array}\right\}
$$

Thus $a+b=2$ or $a+b=-2$, and $a=b$. So $a=1, b=1$ or $a=-1, b=-1$.

Due to the constraints, the only solution is: $a=1, b=1$.