1) To solve the inequality -2X-6+4X^2 > 0, we first rearrange it to the standard form of a quadratic inequality: 4X^2 - 2X - 6 > 0.

Next, we find the roots of the quadratic by setting it equal to 0: 4X^2 - 2X - 6 = 0. Use the quadratic formula to find the roots: X = $2\pm sqrt(4+96)$/8 = $2\pm sqrt(100)$/8 = $2\pm 10$/8.

So, the roots are X = 3/2 and X = -1/2. These divide the number line into three intervals: $-\infty ,-1/2$, $-1/2,3/2$, and $3/2,\infty$.

Now, we test a point in each interval to determine where the inequality holds true. For example, if X = -1, then 4$-1$^2 - 2$-1$ - 6 = 4 + 2 - 6 = 0, which is not greater than 0. Therefore, the inequality is satisfied in the interval $-1/2,3/2$.

Therefore, the solution to the inequality -2X-6+4X^2 > 0 is X ∈ $-1/2,3/2$.

2) To solve the inequality -49 + 36X^2 > 0, we first rearrange it to the standard form of a quadratic inequality: 36X^2 - 49 > 0.

Next, we find the roots of the quadratic by setting it equal to 0: 36X^2 - 49 = 0. Use the difference of squares formula to factor the quadratic: $6X+7$$6X-7$ = 0. Therefore, the roots are X = 7/6 and X = -7/6.

These roots divide the number line into three intervals: $-\infty ,-7/6$, $-7/6,7/6$, and $7/6,\infty$.

Now, we test a point in each interval to determine where the inequality holds true. For example, if X = 0, then 36$0$^2 - 49 = -49, which is less than 0. Therefore, the inequality is satisfied in the intervals $-7/6,7/6$.

Therefore, the solution to the inequality -49 + 36X^2 > 0 is X ∈ $-7/6,7/6$.

3) To solve the inequality -2X-3+5X^2 ≥ 0, we first rearrange it to the standard form of a quadratic inequality: 5X^2 - 2X - 3 ≥ 0.

Next, we find the roots of the quadratic by setting it equal to 0: 5X^2 - 2X - 3 = 0. Use the quadratic formula to find the roots: X = $2\pm sqrt(4+60)$/10 = $2\pm sqrt(64)$/10 = $2\pm 8$/10.

So, the roots are X = 1/5 and X = -3/5. These divide the number line into three intervals: $-\infty ,-3/5$, $-3/5,1/5$, and $1/5,\infty$.

Now, we test a point in each interval to determine where the inequality holds true. For example, if X = 0, then 5$0$^2 - 2$0$ - 3 = -3, which is less than 0. Therefore, the inequality is satisfied in the intervals $-3/5,1/5$.

Therefore, the solution to the inequality -2X-3+5X^2 ≥ 0 is X ∈ $$-3/5,1/5$$.