Step 1: Read the expression.
We must evaluate $\tan\!\left(\cos^{-1}\tfrac{4}{5} + \tan^{-1}\tfrac{2}{3}\right)$.
Step 2: Turn the inverse cosine into an inverse tangent.
Let $\theta = \cos^{-1}\tfrac{4}{5}$, so $\cos\theta = \tfrac{4}{5}$. In a 3-4-5 triangle the opposite side is 3, giving $\tan\theta = \tfrac{3}{4}$, i.e. $\theta = \tan^{-1}\tfrac{3}{4}$.
Step 3: Rewrite the whole expression.
It becomes $\tan\!\left(\tan^{-1}\tfrac{3}{4} + \tan^{-1}\tfrac{2}{3}\right)$.
Step 4: Apply the tangent addition rule.
$\tan(\tan^{-1}X + \tan^{-1}Y) = \dfrac{X+Y}{1-XY}$ with $X = \tfrac{3}{4}$, $Y = \tfrac{2}{3}$.
Step 5: Evaluate numerator and denominator.
Numerator $= \tfrac{3}{4}+\tfrac{2}{3} = \tfrac{9+8}{12} = \tfrac{17}{12}$; denominator $= 1 - \tfrac{3}{4}\cdot\tfrac{2}{3} = 1 - \tfrac{6}{12} = \tfrac{6}{12}$.
Step 6: Divide to finish.
$\dfrac{17/12}{6/12} = \dfrac{17}{6}$, which is option 1 and agrees with the key.
\[ \boxed{\dfrac{17}{6}} \]