Given, the expression is
$5pq(p^2-q^2).$ 
$\text{[Using the identity:}{a}^2-b^2=(a+b)(a-b)]$
$\Rightarrow 5pq(p^2-q^2)=5pq(p+q)(p-q)$
Now dividing the above expression by
$2p(p+q),\text{we get}$
$\frac{5pq(p+q)(p-q)}{2p(p+q)}$
$=\frac{5q(p-q)}{2}$

The given expression is
$(2x{)}^2+5x$

Substituting $x$ = 3 in the above expression we get
$(2\times 3{)}^2+5\times 3$
$6^2+5\times 3$
36 +15  = 51
Hence, the given statemnet is true

$5(x+2y)=5x+5\times 2y=5x+10y$

$\frac{3x^2+1}{3x^2}$
$=\frac{3x^2}{3x^2}+\frac{1}{3x^2}$
$=1+\frac{1}{3x^2}$
Hence, the given statement is false.

$\frac{72}{8}=9\frac{y^4}{y^2}=y^2\frac{z^6}{z^3}=z^3$
Therefore, $\frac{72x^2{y}^4{z}^6}{8x^2{y}^2{z}^3}=9y^2{z}^3$

The given algebraic equation is $x^2-6x-16$.
We can factorize this using the following identity using,
$(x+a)(x+b)=x^2+(a+b)x+ab$
So, we have to first find the factors of 16.
$8\times 2=4\times 4=16$
We can see that, $8-2=6$.
So, we split the expression as shown
$x^2+2x-8x-16$
$=x(x+2)-8(x+2)$
$=(x+2)(x-8)$
Hence $x-8$ is a factor of the given expression.
 

Substituting $x=-3$ in $x^2-5x$ gives 
$(-3{)}^2-5(-3)=9+15=24$