1. From the condition about 15 kg, it follows that the container can contain either only items weighing 5 and 10 kg, or only items weighing 2, 3, and 10 kg. Let $x, y, z, u$ denote the number of items weighing $2, 3, 5$, and 10 kg, respectively.

In the first case, the conditions of the problem give the system $\left\{\begin{array}{l}u=z+5, \\ 5 z+10 u=100,\end{array}\right.$ which has no solutions in natural numbers.

In the second case, for natural numbers $x, y, u$, we obtain the system of equations

$$
\left\{\begin{array} { l } 
{ u = x + y + 5 , } \\
{ 2 x + 3 y + 1 0 u = 1 0 0 , }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ u = x + y + 5 , } \\
{ 1 2 x = 5 0 - 1 3 y , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=y=2 \\
u=9
\end{array}\right.\right.\right.
$$
Answer: 2 items weighing 2 and 3 kg, 9 items weighing 10 kg.

Answer to variant 4-2: 2 weighing 3 and 5 kg, 12 weighing 7 kg.