Should I plot them?

The slope of the tangent to $y = \dfrac 1{2x}+3$ at $x=-1$ is $\dfrac {dy}{dx} \bigg|_{x=-1} = - \dfrac 1{2x^2} \bigg|_{x=-1} = - \dfrac 12$ .

Then $y(-1) = \dfrac 1{2(-1)}+3 = \dfrac 52$ and the equation of the tangent is $\dfrac {y-\frac 52}{x +1} = - \dfrac 12$ , $\implies y = 2 - \dfrac x2$ .

When the tangent meet the other curve $y = 2x^2 + 2$ :

$\begin{aligned} 2 - \frac x2 & = 2x^2+2 \\ 2x^2 + \frac x2 & = 0 \\ 4x^2 + x & = 0 \\ x(4x+1) = 0 \\ \implies x & = - \frac 14 & \small \color{#3D99F6} \text{Since} x < 0 \end{aligned}$

The slope of the curve $y = 2x^2 + 2$ at $x = -\dfrac 14$ is $\dfrac {dy}{dx} \bigg|_{x=-\frac 14} = 4x \bigg|_{x=-\frac 14} = - 1$ .

The angle between the tangent and the curve is $\tan^{-1} 1 - \tan^{-1} \dfrac 12 \approx 45^\circ - 26^\circ = \boxed{19}^\circ$