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(A) To answer these questions, we first solve the indicated inequality $\left(x^2>x^4\right)$ for all real numbers $x\in\mathbb{R}$ .

$x^2-x^4>0\to\left.x^2\cdot(1-x^2)>0\right|\cdot(-1)\to\\[0.3cm] x^2\cdot\left(x^2-1\right)\equiv\boxed{x^2\cdot(x-1)(x+1)<0}$

Conclusion,

$\boxed{x\in(-1;0)\cup(0;1)}\\[0.3cm] \text{As you can see, the solutions of this inequality are NOT integers}$

Moving on to the answers to these questions :

1. $P(0) : 0^2>0^4-\boxed{\text{FALSE}}$

2. $P(-1) : (-1)^2=1>(-1)^4=1-\boxed{\text{FALSE}}$

3. $P(1) : 1^2=1>1^4=1-\boxed{\text{FALSE}}$

4. $P(2) : 2^2=4>2^4=16-\boxed{\text{FALSE}}$

5. $\exists xP(x) : x^2>x^4-\boxed{\text{FALSE}}$

6. $\forall xP(x) : x^2 >x^4-\boxed{\text{FALSE}}$

(B) Let $P(x)$ be the statement $(x>1)$ and $Q(x)$ is $\left(x-1>1\right)$. Then, the sentence " for every real number x, if $x>1$ then $x-1>1$ " has the form

$\boxed{\forall x\in\mathbb{R}\left(P(x)\to Q(x)\right)-\text{FALSE}}$

For example,

$x=1.1\to\left\{\begin{array}{l} P(1.1)=1.1>1-\text{TRUE}\\ Q(1.1)=1.1-1=0.1>1-\text{FALSE} \end{array}\right. \\[0.3cm] x=1.2\to\left\{\begin{array}{l} P(1.2)=1.2>1-\text{TRUE}\\ Q(1.2)=1.2-1=0.2>1-\text{FALSE} \end{array}\right. \\[0.3cm] x=1.3\to\left\{\begin{array}{l} P(1.3)=1.3>1-\text{TRUE}\\ Q(1.3)=1.3-1=0.3>1-\text{FALSE} \end{array}\right. \\[0.3cm]$

Let $P(x)$ be the statement $(x^2\le0)$. Then, the sentence " for some real number $x$ , $x^2\le0$ " has the form

$\boxed{\exists xP(x)-\text{TRUE}}$

For example,

$x=0\to 0^2\le0-\text{TRUE} \\[0.3cm] x=1.2\to 1.2^2\equiv1.44\le0-\text{FALSE}\\[0.3cm] x=-2.5\to(-2.5)^2\equiv6.25\le0-\text{FALSE}$

(C)

1.No one in this class is wearing pants and a guitarist.

Let:

Domain of $x$ is all persons

$A(x) : x$ is wearing pants

$B(x) : x$ is a guitarist

$C(x) :$ belongs to the class

$\boxed{\forall x\in D,\left(C(x)\to\left(A(x)\land B(x)\right)\right)}$

2. No one in this class is wearing pants and a guitarist.

Let:

Domain of $x$ is persons in this class

$A(x):x$ is wearing pants

$B(x):x$ is a guitarist

$\boxed{\forall x\left(\neg A(x)\land\neg B(x)\right)}$

3. There is a student at your school who knows C++ but who doesn’t

know Java.

Let:

Domain: all students at your school

$C(x):x$ knows C++

$J(x):x$ knows Java

$\boxed{\exists x\left(C(x)\land\neg J(x)\right)}$

Q.E.D.