Here, 
$-7-6=-13,$
$-5+1=-4,$
$2-1=1,$
$8-9=-1$
$\therefore$ The second statement is false.

RHS = (-4 $\times$ 2) + 10 = 2

LHS = (6 - 7) $\times$ -2 = 2
$\therefore$ (-4 $\times$ 2) + 10 = (6 - 7) $\times$ -2

Any integer multiplied by '1' results in the same integer. So, the multiplicative identity for integers is 1.
1 is also the first natural number so we can say that even the first natural number is also a multiplicative identity for integers.

The reciprocal of $\frac{1}{6}$ is 6

If 'a' and 'b' are any two integers, by distributive property of multiplication over addition, 

a × (b + c) = (a × b) + (a × c)

(i) All the negative integers are less than 0. 

(ii) Multiplicative identity for integers is same as that of  whole numbers i.e., 1.

(iii) Absolute value of a non-negative integer is equal to the integer itself and the absolute value of a negative integer is always greater than the integer.

E.g. absolute value of 6 is 6; absolute value of - 6 is 6.

(iv) Every negative integer is less than any natural number.

The predecessor of a number is found by subtracting $1$ from it.

Therefore, predecessor of $-9$ is given as,

$-9-1=-10$

When even number of negative integers are multiplied, the product will be a positive integer.

(-1) x (-1) x (-1) x (-1) x ... 80 times

= (-1 ... 80 times)[1 x 1 x 1 x 1 x ... 80 times]

= (+) (1 x 1 x 1 x 1 x ... 80 times)

= 1 x 1 x 1 x 1 x ... 80 times
= 1

Since the number 0 is to the right of -1 on the number line, 0 is larger than -1. So, the statement is FALSE.