For simplicity, let's assume that $5$ divides $n$
and that $n/2=n-5k$ for some integer $k>0$.
$$ \begin{align*}
T(n) &= T(n-5) + cn^2 \\
T(n) &= cn^2 + c(n - 5)^2 + c(n - 10)^2 + c(n - 15)^2 + ... + c5^2 + c0^2 \\
&= c(n^2 + (n - 5)^2 + (n - 10)^2 + (n - 15)^2 + ... + 5^2 + 0^2) \\
&\ge c(n^2 + (n - 5)^2 + (n - 10)^2 + (n / 2)^2)
\ge c(n / 2)(1 / 5)(n / 2)^2) \\
&= cn^3 / 40
= (c / 40)n^3 \\
&= \Omega(n^3) \\
T(n) &= cn^2 + c(n - 5)^2 + c(n - 10)^2 + c(n - 15)^2 + ... + c5^2 + c0^2 \\
&\le c(n / 5)n^2
\le cn^3 \\
&= O(n^3) \\
\end{align*} $$
We conclude that $T(n) = \Theta(n^3)$.
Let's assume for simplicity that $n/2 = n-2k$ for an integer $k>1$.
$$ \begin{align*}
T(n) &= T(n-2) + \log n
= \log n + \log(n - 2) + \log(n - 4) + ... + \log(4) \\
&\ge \log n + \log(n - 2) + \log(n - 4) + ... + \log(n / 2)
\ge (n / 2)log(n / 2) \\
&= \Omega(n \log n) \\
T(n) &= T(n-2) + \log n
= \log n + \log(n - 2) + \log(n - 4) + ... + \log(4) \\
&\le (n / 2) \log n \\
&= O(n \log n) \\
\end{align*} $$
We conclude that $T(n)=\Theta(n \log n)$.