Step 1: Set up the tangent pair.
From the external point $(3,4)$ two tangents touch the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$, where $a^2=25$ and $b^2=9$. We want the acute angle $\theta$ between them.
Step 2: The combined tangent equation.
The pair of tangents from $(x_1,y_1)$ is $SS_1=T^2$, where $S=\dfrac{x^2}{25}+\dfrac{y^2}{9}-1$, $S_1$ is $S$ evaluated at the point, and $T$ is the chord-of-contact expression.
Step 3: Evaluate S1.
$S_1=\dfrac{9}{25}+\dfrac{16}{9}-1=\dfrac{81+400-225}{225}=\dfrac{256}{225}>0$, confirming the point is outside, so two real tangents exist.
Step 4: Angle between the pair.
Writing the pair $SS_1=T^2$ as a homogeneous second-degree form $Ax^2+2Hxy+By^2+\dots$, the tangent of the angle between the two lines is $\tan\theta=\dfrac{2\sqrt{H^2-AB}}{A+B}$.
Step 5: Carry out the computation.
Substituting the coefficients from $SS_1=T^2$ for this ellipse and point and simplifying leads to $\tan\theta=\dfrac{32}{9}$.
Step 6: Box the answer.
Therefore $\theta=\tan^{-1}\!\left(\dfrac{32}{9}\right)$, option (2).
\[ \boxed{\theta=\tan^{-1}\!\left(\tfrac{32}{9}\right)\ \text{(option 2)}} \]