1.1.1 The equation $x(x-7)=0$ has two solutions: $x=0$ and $x=7$. This can be seen by factoring the left-hand side of the equation as $x(x-7)=x\cdot x - 7\cdot x = 0$, which yields the quadratic equation $x^2 - 7x = 0$. The solutions to this quadratic equation are $x=0$ and $x=7$. Therefore, the solutions to the original equation are $x=0$ and $x=7$.

1.1.2 The equation $x^2-6x+2=0$ has two solutions, which can be found using the quadratic formula:

$x = \frac{6 \pm \sqrt{6^2-4\cdot 2}}{2\cdot 2} = \frac{6 \pm \sqrt{36-8}}{4} = \frac{6 \pm \sqrt{28}}{4} = \frac{6 \pm 2\sqrt{7}}{4}$

Therefore, the solutions to the original equation are $x=\frac{6 + 2\sqrt{7}}{4} \approx 2.91$ and $x=\frac{6 - 2\sqrt{7}}{4} \approx 0.09$.

1.1.3 The equation $\sqrt{x-1}+1=x$ can be rewritten as $\sqrt{x-1} = x-1$, which means that $x-1$ must be non-negative. This implies that $x \ge 1$. Substituting $x=1$ into the original equation yields $\sqrt{1-1}+1=1$, which is not a valid solution because the square root of a negative number is undefined. Therefore, the only solution to the equation is $x=1$.

1.1.4 The equation $3^{x+3}-3^{x+2}=486$ can be rewritten as $3^{x+3} = 3^{x+2}+486$, which implies that $3^{x+3}$ is equal to the sum of two integer powers of 3. This is only possible if $x+3$ is a multiple of 3, which means that $x=-3+3k$ for some integer value of $k$. Substituting this expression for $x$ into the original equation yields $3^{-3+3k} = 3^{-2+3k}+486$, which simplifies to $3^{3k} = 3^{3k-1}+486$. Since $3^{3k}$ and $3^{3k-1}$ are both integer powers of 3, this equation is only possible if $3^{3k-1}$ is divisible by 487.

We can find the values of $k$ that satisfy this condition by testing each possible value of $3^{3k-1}$ modulo 487. If $3^{3k-1}$ is divisible by 487, then $3^{3k-1} \equiv 0 \pmod{487}$, which means that $3k-1$ must be a multiple of 487. We can test this condition by checking the values of $3k$ modulo 487 for each possible value of $k$: