To compute time complexity, we can ignore constant part, because it won't make a difference.
and I think it's 0 not d? It's a base case.
So we can re-write both expressions and then can apply Master's theorem:
$1. T(n)=3T(\frac{n}{2}) + n^{\frac{1}{2}} $ if n>1
a=3, b=2, f(n)=$n^{\frac{1}{2}} $
$n^{\log _{2}^{3}} > n^{\frac{1}{2}}$
Hence, T.C will be $ \Theta (n^{\log _{2}^{3}}) $
$2. T(n)=3T(\frac{n}{2}) + n^{2} $ if n>1
$n^{2} > n^{\log _{2}^{3}} $
Hence, T.C will be $\Theta(n^{2})$