(1) Analysis

This question examines the issue of the inequality always holding true, paying attention to the characteristics of the corresponding quadratic function graph, reflecting the mathematical idea of equivalent transformation, and is considered a medium-level question.

Transform the condition into $2x^{2}-3ax+9\geqslant 0$ always holding true, through $\Delta =9a^{2}-72\leqslant 0$, thereby solving the range of values for the real number $a$.

Solution

Since the proposition "There exists $x \in \mathbb{R}$, such that $2x^{2}-3ax+9 0$, hence $-(2^{x}-1)^{2}+1\leqslant 1$,

Therefore $m\leqslant 1$,
Since $\neg p$ is a false proposition, then $p$ is a true proposition,
Therefore $m\leqslant 1$,
Thus, the answer is $\boxed{m\leqslant 1}$.

(4) Analysis

This question mainly examines the calculation of definite integrals, with the key being finding the antiderivative, considered a basic question.

Calculate according to the rules of definite integration.

Solution

$\int_{0}^{\pi}(x+\cos x)dx=\left( \dfrac{1}{2}x^{2}+\sin x\right)\bigg|_{0}^{\pi}= \dfrac{\pi^{2}}{2}$.

Therefore, the answer is $\boxed{\dfrac{\pi^{2}}{2}}$.