1. Consider the vectors $\vec{a}=(3, x), \vec{b}=(\sqrt{x-2}, 4)$, then $\vec{a} \cdot \vec{b} \leq|\vec{a}||\vec{b}|$, which is equivalent to

$$
3 \sqrt{x-2}+4 x \leq \sqrt{\left(9+x^{2}\right)(x+14)}
$$

Equality is possible if and only if the vectors are collinear

$$
\frac{\sqrt{x-2}}{3}=\frac{4}{x}>0 \Leftrightarrow x \sqrt{x-2}=12
$$

The left side of the considered equation is a monotonically increasing function on the domain of definition, so the equation has no more than one root. By trial, we find $x=6$.
Answer: 1. 6. 2.6 3.74. 7.

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