The sum of squares of intercepts on the axes cut off by the tangents to the curve \(\mathrm{x}^{(2 / 3)}+\mathrm{y}(2 / 3)=\mathrm{a}^{(2 / 3)}(\mathrm{a}>0)\) at \([(\mathrm{a} / 8),(\mathrm{a} / 8)]\) is \(2 .\) Thus a has the value. (a) 1 (b) 2 (c) 4 (d) 8

The equation of the given curve is \(x^\frac{2}{3} + y^\frac{2}{3}= a^\frac{2}{3}\). Now, to find the equation of tangents passing through the point \(\left( \frac{a}{8},\frac{a}{8} \right)\), we differentiate the curve equation with respect to \(x\), which gives us the slope of the tangent. \[ \frac{d}{dx} \left( x^\frac{2}{3} + y^\frac{2}{3} \right) = 0\] \[\frac{2}{3}x^{\frac{-1}{3}} + \frac{2}{3}y^{\frac{-1}{3}} \frac{dy}{dx} = 0\] Now, run the given point or coordinates \(\left( \frac{a}{8},\frac{a}{8}\right)\) through the equation. \[\frac{2}{3}\left(\frac{a}{8}\right)^{-\frac{1}{3}} + \frac{2}{3}\left(\frac{a}{8}\right)^{-\frac{1}{3}} \frac{dy}{dx} = 0\] Next, solve for the slope: \[\frac{dy}{dx} = -\frac{2x^\frac{-1}{3}}{2y^\frac{-1}{3}} = -\frac{x^\frac{-1}{3}}{y^\frac{-1}{3}}\] At the point \(\left( \frac{a}{8},\frac{a}{8} \right)\), the slope of the tangent is \(-1\). With the slope and the point on the tangent, we can now use the point-slope form to find the equation of the tangent: \[y - \frac{a}{8} = -1 \cdot \left(x - \frac{a}{8}\right)\]

Now, we'll find the x-intercept and y-intercept of the tangent line. For x-intercept, set y = 0: \[0 - \frac{a}{8} = -1 \cdot \left(x - \frac{a}{8}\right)\] For y-intercept, set x = 0: \[y - \frac{a}{8} = -1 \cdot \left(0 - \frac{a}{8}\right)\] Solving these equations will give us the intercept points

Now, we will calculate the sum of squares of intercepts and set it equal to 2: \[\left(\frac{a^2}{64}\right) + \left(\frac{a^2}{64}\right) = 2\] Solve the equation for \(a\) and find its value.

\[\frac{a^2}{32} = 2\] \[a^2 = 64\] \[a = \pm 8\] Since given that \(a > 0\), the correct answer is: \(a = 8\) Hence, the value of \(a\) is 8 (Option d).