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$N$ divided by 3 leaves a remainder of 2 means $N = 3k+2$ for some integer $k$.  $N$ divided by 4 leaves a remainder of 2 means $N = 4\ell+2$ for some integer $\ell$.   It follows that $3k+2=4\ell+2$, so that $3k=4\ell$.  This tells us that $\ell = 3m$ for some integer $m$.   Replacing $\ell$ by $3m$ in the equation $N = 4\ell+2$ should give you the answer.

By Vieta's Theorem the equation $3x^2+5x=7=0$ gives you the values of $u+v$ and $uv$.   Noticing that $u^2+v^2 = (u+v)^2 - 2uv$ should tell you what to do.

From the assumptions it follows that 2, 3, and 5 are all divisors of $n$.   Also $n$ should not be divisible by any other primes because if it were then dividing $n$ by the product of all these other primes would result in still another integer satisfying all the given assumptions.

Thus we conclude that $n= 2^a\cdot 3^b\cdot 5^c$ for some nonnegative integer $a$, $b$, and $c$.

The assumption $2n$ being a perfect square then implies $2 \mid a+1$, $2 \mid b$, and $2 \mid c$.

The assumption $3n$ being a perfect cube then implies $3 \mid a$, $3 \mid b+1$, and $3 \mid c$.

The assumption $5n$ being a perfect fifth power then implies $5 \mid a$, $5 \mid b$, and $5 \mid c+1$.

So $a$ has to be a multiple of ${\rm lcm}(3, 5)$ such that $a+1$ is even,
$b$ has to be a multiple of ${\rm lcm}(2, 5)$ such that $b+1$ is divisible by 3,

and $c$ has to be a multiple of ${\rm lcm}(2, 3)$ such that $c+1$ is divisible by 5.

The least numbers satisfying these conditions are $a=15$, $b=20$, and $c=24$.  

Thus $n=2^{15}\cdot 3^{20}\cdot 5^{24}$.

Write $t = n+x$ where $n$ is $\lfloor t\rfloor$ and $x$ is the fractional part ($0\le x<1$).  Given that $\lfloor t\rfloor = 2t+5$.  We have
\begin{eqnarray*}
    \lfloor n+x \rfloor = 2(n+x)+5 \\
    n = 2n+2x+5 \\
    n = -5-2x
\end{eqnarray*}
Since $n$ is an integer, $2x$ must also be an integer.   And since $0\le x<1$,  we have $0\le 2x < 2$, so $x$ is either 0 or 0.5.  When $x=0$, we have $n=-5-2\cdot 0=-5$.  When $x=0.5$, we have $n=-5-2\cdot 0.5=-6$.  Thus, the possible values of $t$ in increasing order are $-6+.5=-5.5$, $-5+0=-5$.

For brevity, all congruences below are taken modulo $n$.  $(ab)^{-1}\equiv 2$ means $2(ab)\equiv1$.  So we have $(2a)b\equiv a(2b)\equiv1$, meaning both $a$ and $b$ are invertible with $a^{-1}= 2b\bmod n$ and $b^{-1}=2a\bmod n$.  Thus,
\begin{eqnarray*}
    (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& (a+b)^{-1}(2b+2a) \\
    &\equiv& ((a+b)^{-1}(a+b))2 \\
    &\equiv& 1\cdot 2 \\
    &\equiv& 2.
\end{eqnarray*}

We have $3x+7 \equiv 2 \pmod{16}$, so $3x \equiv -5 \equiv 11 \pmod{16}$.
This solves to $x \equiv 9 \pmod{16}$.

Thus, $2x+11 \equiv 2\cdot 9+11\equiv 13 \pmod{16}$.

I claim that the $n=m=15$, so $n-m=0$.  To see that $15!$ is a common multiple of 4125 and 2816000, just check that $15!$ is a multiple of both $4125=3\cdot 5^3\cdot 11$  and $2816000=2^{11}\cdot 5^3\cdot 11$.  To see that $15!$ is the smallest common multiple of 4125 and 2816000, note that $14!$ is not a common multiple of 4125 and 2816000 (because $5^3$ does not divide $14!$ so it can't divide any $x!$ where $x \le 14$).

The term $a*b$ is nonnegative whenever both $a$ and $b$ have the same sign. It is negative whenever $a$ is positive and $b$ is negative.   Since $a$, $b$ are integers with $a = b+10$, these are all the possible ways that $ab$ can be negative:
\begin{eqnarray*}
    a = 1, b = -9, ab = -9 \\
    a = 2, b = -8, ab = -16 \\
    a = 3, b = -7, ab = -21 \\
    a = 4, b = -6, ab = -24 \\
    a = 5, b = -5, ab = -25 \\
    a = 6, b = -4, ab = -24 \\
    a = 7, b = -3, ab = -21 \\
    a = 8, b = -2, ab = -16 \\
    a = 9, b = -1, ab = -9
\end{eqnarray*}

So the minimum value of $ab$ is $-25$.

By definition $33^{-1} \equiv 77 \pmod{508}$ means $33\cdot 77 \equiv 1 \pmod{508}$.  Thus $11\cdot 3\cdot 77 \equiv 1 \pmod{508}$.  Hence, $11^{-1} \equiv  3\cdot 77 \equiv 231 \pmod{508}$.

We have $y = \frac{x - 3}{x + 1} = 1 - \frac{4}{x+1}$.  So $\frac{4}{x+1} = 1-y$.  Since $x\ne -1$ we can multiply $x+1$ to both sides to get $(x+1)(1-y)=4$, or equivalently, $(x+1)(y-1)=-4$, i.e., $a=1$, $b=-1$ and $c=-4$.