 Solution

For the first recurrence relation, the master theorem compares $n^2$ with $n^{{\log }_{2}6}$. Since $2<{\log }_{2}6$, the recursive leaves dominate and the answer is $\Theta (n^{{\log }_{2}6})$. More explicitly, the work at depth $j$ is $n^2(6/4{)}^j$, a growing geometric series that stops at depth ${\log }_{2}n-O(1)$, where its sum has that order.

For the second recurrence relation, each step subtracts two and adds one until reaching a base case. For integral $n>4$ there are $\lceil (n-4)/2\rceil$ steps, so $T(n)=1+\lceil (n-4)/2\rceil$ and the answer is $\Theta (n)$.

The third set of printed requirements is inconsistent. The specified base values and its recurrence relation give $T(n)=1$ identically, as strong induction immediately verifies: $4\cdot 1-4\cdot 1+1=1$. In particular $T(8)=1$, while the extra condition demands $T(8)>2T(4)=2$. Thus no function satisfies all the printed conditions. Dropping that extra inequality gives $\Theta (1)$.

The likely intended growth calculation can nevertheless be made precise, without treating it as a solution of the contradictory data. On dyadic arguments put $U_j=T(2^j)$. Its equation has the general formbecause $U_j-1$ has characteristic equation $(r-2{)}^2=0$. The printed values $U_1=U_2=1$ force $A=B=0$. With suitably different initial values, the condition $U_j>2U_{j-1}$ for arbitrarily large $j$ instead requires $B{2}^j-1>0$, hence $B>0$ and $U_j=\Theta (j{2}^j)$, or $T(n)=\Theta (n\log n)$. That last order describes corrected initial data, not the paper as printed. All three bullets belong to this single marked part.