$3 \sqrt {1.02}+4 \sqrt {1.02}$

Step 1:

Let $y=x^{\large\frac{1}{3}}+x^{\large\frac{1}{4}}$

$dy=\bigg(\large\frac{1}{3}x^{\Large\frac{-2}{3}}+\large\frac{1}{4}x^{\Large\frac{-3}{4}}\bigg)$$dx$

$\quad=\bigg(\large\frac{1}{3x^{\Large\frac{-2}{3}}}+\large\frac{1}{4x^{\Large\frac{-3}{4}}}\bigg)$$dx$

Step 2:

Let $x=1$ and $dx \approx \Delta x=0.02$

Then $f(1)=1^{\large\frac{1}{3}}+1^{\large\frac{1}{4}}$$=2$

Also $dy=\bigg(\large\frac{1}{3}+\frac{1}{4}\bigg)$$(0.02)$

$\qquad=\large\frac{7}{12} $$(0.02)$

$\qquad=0.0117$

$\therefore f(1.02)=f(1)+dy$

$\qquad=2+0.0117=2.0117$