I'll do the first two parts to get you started.
=====================================================
Part (i)
We can show that the operation [tex]\nabla[/tex] is not commutative by picking two random and different real numbers for a,b. In other words, I'm using a counter-example.
Let's say we go for a = 1 and b = 2
[tex]a \nabla b = \frac{3a+b}{5} - 1\\\\1 \nabla 2 = \frac{3*1+2}{5} - 1\\\\1 \nabla 2 = \frac{5}{5} - 1\\\\1 \nabla 2 = 1 - 1\\\\1 \nabla 2 = 0\\\\[/tex]
Now let's swap the values. We'll try a = 2 and b = 1
[tex]a \nabla b = \frac{3a+b}{5} - 1\\\\2 \nabla 1 = \frac{3*2+1}{5} - 1\\\\2 \nabla 1 = \frac{7}{5} - 1\\\\2 \nabla 1 = 1.4 - 1\\\\2 \nabla 1 = 0.4\\\\[/tex]
We can see that [tex]1 \nabla 2[/tex] and [tex]2 \nabla 1[/tex] are not the same value. In general, [tex]a \nabla b \ne b \nabla a[/tex] Therefore, the operation [tex]\nabla[/tex] is not commutative.
The only time [tex]a \nabla b = b \nabla a\\\\[/tex] is true is when [tex]a = b[/tex], since,
[tex]\frac{3a+b}{5}-1 = \frac{3b+a}{5}-1\\\\\frac{3a+a}{5}-1 = \frac{3a+a}{5}-1\\\\\frac{4a}{5}-1 = \frac{4a}{5}-1\\\\[/tex]
It's when [tex]a \ne b[/tex] is where the operation becomes noncommutative.
Another way to arrive at the [tex]a = b[/tex] condition is to solve the original equation for either 'a' or b.
=====================================================
Part (ii)
Using the previous part as inspiration, we'll do a counter-example to show that the operation is not associative. Pick 3 random values for a,b,c. Here are the values I'll pick.
Then,
[tex]d = b \nabla c = \frac{3b+c}{5}-1 = \frac{3*2+3}{5}-1 = 0.8\\\\a \nabla (b \nabla c) = a \nabla d = \frac{3a+d}{5}-1 = \frac{3*1+0.8}{5}-1 = -0.24\\[/tex]
Next, we can say:
[tex]d = a \nabla b = \frac{3a+b}{5}-1 = \frac{3*1+2}{5}-1 = 0\\\\(a \nabla b) \nabla c = d \ \nabla c = \frac{3d+c}{5}-1 = \frac{3*0+3}{5}-1 = -0.4[/tex]
Let's compare the two outputs:
[tex]a \nabla (b \nabla c) = -0.24\\\\(a \nabla b) \nabla c = -0.4[/tex]
They don't match up, so [tex]a \nabla (b \nabla c) \ne (a \nabla b) \nabla c[/tex] when (a,b,c) = (1,2,3).
In general, the operation is not associative in R.
