曲线方程为 $y = 2x^2 + 5$。

1. Substitute the expression for $y$ into the given inequality

$2x^2 + 5 > 3x^2$

2. Rearrange the inequality into standard quadratic form

$x^2 - 5 < 0$

3. Solve the quadratic inequality

$(x - \sqrt{5})(x + \sqrt{5}) < 0$

$-\sqrt{5} < x < \sqrt{5}$

4. Represent the solution on a number line

Draw a number line with open circles at $-\sqrt{5}$ and $\sqrt{5}$, and a line segment connecting them to indicate the values between them.

1. Substitute the equation of the line into the equation of the curve

From the line equation, $y = kx + 3$.
Substitute into the curve equation:
$kx + 3 = 2x^2 + 5$

2. Rearrange into a standard quadratic equation

$2x^2 - kx + 2 = 0$

3. Apply the condition for a tangent (discriminant equals zero)

For the line to be a tangent, the discriminant $b^2 - 4ac = 0$.
$(-k)^2 - 4(2)(2) = 0$
$k^2 - 16 = 0$

4. Solve for $k$

$k^2 = 16$
$k = \pm 4$

1. Determine the value of $k$ and set up the equation for the x-coordinate of $P$

From part (b), the condition for the line to be a tangent gives $k^2 - 16 = 0$, so $k = \pm 4$.
Given $k > 0$, we have $k = 4$.
The intersection equation is $2x^2 - kx + 2 = 0$.
Substituting $k = 4$ gives $2x^2 - 4x + 2 = 0$.

2. Solve the quadratic equation to find the x-coordinate of $P$

$2(x^2 - 2x + 1) = 0$
$2(x - 1)^2 = 0$
$x = 1$

3. Find the corresponding y-coordinate of $P$

Substitute $x = 1$ into the curve equation $y = 2x^2 + 5$:
$y = 2(1)^2 + 5 = 7$
The coordinates of $P$ are $(1, 7)$.