1. For the given expression $V$, we have

$$
V=a(b+d)+c(b+d)=(a+c)(b+d).
$$

Similarly, we can modify both given conditions:

$$
2(a+c)-5(b+d)=4 \quad \text { and } \quad 3(a+c)+4(b+d)=6
$$

If we choose the substitution $m=a+c$ and $n=b+d$, we get the solution of the system (1) as $m=2$ and $n=0$. For the given expression, we then have $V=m n=0$.

Conclusion. Under the given conditions, the expression $V$ takes only the value 0.

Alternative solution. We can interpret the conditions of the problem as a system of equations with unknowns $a, b$ and parameters $c, d$. By solving this system (using the addition or substitution method), we find $a=2-c, b=-d$ (where $c, d \in \mathbb{R}$), and substituting into the expression $V$ gives

$$
V=(2-c)(-d)-d c+c d+d(2-c)=0.
$$

For a complete solution, award 6 points. For the first solution, award 2 points for factoring the expression $V$ into a product, 2 points for transforming the conditions into the system (1), 1 point for solving the system, and 1 point for calculating the value of $V$.
1. Determine the values that the expression $V=a b+b c+c d+d a$ can take, given that the real numbers $a, b, c, d$ satisfy the following conditions:

$$
\begin{aligned}
& 2 a-5 b+2 c-5 d=4, \\
& 3 a+4 b+3 c+4 d=6 .
\end{aligned}
$$