In both functions $n$ denotes the size of the input and it is assumed to be non negative. For the base cases we have $T_{1}(1) = c_{1}$ and $T_{2}(1) = c_{2}$ where $c_{1}$ and $c_{2}$ are positive constants.
Analysis of $T_{1}(n) = 3T_{1}(\frac{n}{2}) + n$
$T_{1}$ implies a tree such that every internal node has three children. When we go from one node to a direct child the size of the input decreases by two, so the height of the tree is $\log_{2}(n)$ plus or minus one some constant. We will work with $\log_{2}(n)$ to make the math simpler.
Assuming a zero numbering for the levels of the tree, the total work performed in level $i$ is $3^{i}\frac{n}{2^{i}}$ so we have that $T_{1}(n) = \sum_{i=0}^{\log_{2}(n)-1} 3^{i}\frac{n}{2^{i}} = n \sum_{i=0}^{\log_{2}(n)-1} (\frac{3}{2})^{i} = n\frac{{(\frac{3}{2})}^{\log_{2}(n)}-1}{\frac{3}{2}-1} = \frac{n({(\frac{3}{2})}^{\log_{3/2}(n)})^{\frac{1}{\log_{3/2}{(2)}}}-n}{0.5}=\frac{nn^{\frac{1}{\log_{3/2}{2}}}-n}{0.5} = \frac{n^{1.5849}-n}{0.5} = 2n^{1.5849}-2n=\Theta(n^{1.5849})$
Analysis of $T_{2}(n) = 3T_{2}(\frac{n}{2}) + \log_{2}(n)$
Now every time we are about to split a node, we perform a logarithmic amount of work, so intuition suggests that the overall running time will be asymptotically less, since the same relationship applies for $\log(n)$ and $n$. Let us follow a similar mathematical approach here.
We have
$T_{2}(n) = \sum_{i=0}^{\log_{2}(n)-1} 3^{i}\log_{2}(\frac{n}{2^{i}}) =\sum_{i=0}^{\log_{2}(n)-1} 3^{i}\log_{2}(n) - \sum_{i=0}^{\log_{2}(n)-1} 3^{i}\log_{2}(2^{i}) = \sum_{i=0}^{\log_{2}(n)-1} 3^{i}\log_{2}(n) - \sum_{i=0}^{\log_{2}(n)-1} i3^{i}$
Let $A = \sum_{i=0}^{\log_{2}(n)-1} 3^{i}\log_{2}(n)$ and $B = \sum_{i=0}^{\log_{2}(n)-1} i3^{i}$
$A = \log_{2}(n)\sum_{i=0}^{\log_{2}(n)-1} 3^{i}$. From the geometric series we know that $\sum_{i=0}^{N-1}x^{i} = \frac{x^{N}-1}{x-1}$ so we have that $A = \log_{2}(n)\frac{3^{\log_{2}(n)}-1}{2} = \log_{2}(n)\frac{(3^{\log_{3}(n)})^{\frac{1}{\log_{3}(2)}}-1}{2} = \frac{n^{1.5849}\log_{2}(n)-\log_{2}(n)}{2}$
To find $B$ we need a formula for $\sum_{i=0}^{N-1}ix^{i}$. Let $S = \sum_{i=0}^{N-1}ix^{i} = 1x^{1} + 2x^{2} + 3x^{3} + ... + (N-1)x^{N-1}$
$-xS = -x^{2} - 2x^{3} - 3x^{4} - ... - (N-1)x^{N}$. Combining both we get
$S - xS = x^{1} + x^{2} + x^{3} + ... + x^{N-1} - (N-1)x^{N}$.
Let $S' = x^{1} + x^{2} + ... + x^{N-1}$
$-xS' = -x^{2} - x^{3} - ... - x^{N-1} - x^{N}$
So $S'(1-x) = x - x^{N}$
Combining everything we get $S = \frac{(N-1)x^{N} - \frac{x^{N}-x}{x-1}}{x-1}$.
To find $B$ we need to set $x = 3$ and $N=\log_{2}(n)$. So we have the following:
$B = \frac{(\log_{2}(n)-1)3^{\log_{2}(n)} - \frac{3^{\log_{2}(n)}-3}{3-1}}{3-1} = \frac{\frac{2(\log_{2}(n)-1)3^{\log_{2}(n)}}{2} - \frac{n^{1.5849}-3}{2}}{2} = \frac{2(\log_{2}(n)-1)n^{1.5849} - n^{1.5849}+3}{4} = \frac{2n^{1.5849}\log_{2}(n)-2n^{1.5849}-n^{1.5849}+3}{4}$
Finally we have
$T_{2}(n) = A - B = \frac{n^{1.5849}\log_{2}(n)-\log_{2}(n)}{2} - \frac{2n^{1.5849}\log_{2}(n)-2n^{1.5849}-n^{1.5849}+3}{4} = \frac{2n^{1.5849}\log_{2}(n)-2\log_{2}(n)-2n^{1.5849}\log_{2}(n)+2n^{1.5849}+n^{1.5849}-3}{4} = \frac{3n^{1.5849}-2\log_{2}(n)-3}{4} = \Theta(n^{1.5849})$