a)Let $\alpha=\sqrt[p]{a},\beta=\sqrt[p]{b}$.Now suppose that $f,g$ are the minimal polynomials of $\alpha,\beta$ in $\mathbb{Q}(\alpha+\beta)$.We are going to show $f,g$ are linear.We have $g(\alpha+\beta-x),f(\alpha+\beta-x)\in\mathbb{Q}(\alpha+\beta)$.By the definition of minimal polynomials,we have $f(x)|g(\alpha+\beta-x)$ and similarly $g(x)|f(\alpha+\beta-x)$.So we have $f(x)=g(\alpha+\beta-x)$.By the way,we have $f(x)|x^p-a,g(x)|x^p-b$.So all root of $f,g$ are in form $\varepsilon^i\alpha,\varepsilon^i\beta$ respectively where $\varepsilon^i$'s are pth roots of unity.So if $deg (f)\geq2$ ,then $f$ should have another root like $\varepsilon^n\alpha$ for some integer $n$.So $g(\alpha+\beta-\varepsilon^n\alpha)=0$.So $\alpha+\beta-\varepsilon^i\alpha=\varepsilon^m\beta$ for some integer $m$.You can easily check that this is impossible except the case $p|m,n$.So $deg(f)=1$.This is efficient to prove $\alpha\in\mathbb{Q}(\alpha+\beta)$.And now we can say $\beta\in\mathbb{Q}(\alpha+\beta)$.So $\mathbb{Q}(\alpha,\beta)\subset\mathbb{Q}(\alpha+\beta)$.The other side is obvious.