Finding: $\gcd\left(2^{200}-2^{100},2^{200}+2^{101}\right)$ and $\gcd\left(3^{202}-3^{101},3^{202}+3^{102}\right)$

Yes, your answers are correct and your use of $t$ was well judged. You might like to consider an arguably neater way of writing out the proof:-

Line 1 as yours, then

$$t=2^{100}\equiv(-1)^{100}\equiv1\pmod{3}$$

$\implies t\cdot\gcd\left(t-1,t+2\right)=t\cdot\gcd\left(t-1,t+2-(t-1)\right)=t\cdot\gcd\left(t-1,3\right)=3t$

That works. Simpler by Euclidean algorithm: $\ (x, \color{#0a0}y) = (x, \color{#0a0}{\bar y)}\ $ if $\ \color{#0a0}{y\equiv \bar y}\pmod{\!x},\ $ so

$$\begin{align} (a^{2n}-a^n,\, a^{2n}+a^{n+1})\, &=\, a^n(\ \ a^n\,-\,1,\,\color{#0a0}{a^n}+a)\\[.2em] &=\, a^n(\ \ \color{#c00}a^n\,-\,1,\ \color{#0a0}1\,+\,a)\ \ \ {\rm by}\,\ \ \color{#0a0}{a^n\equiv 1}\!\!\!\pmod{a^n-1\!}\\[.2em] &=\, a^n({\small{(\color{#c00}{-1})}}^n\!-\!1,\ 1\,+\,a)\ \ \ {\rm by}\ \ \color{#c00}{a\equiv -1}\!\!\!\pmod{1+a}\\ \end{align}$$

So for the OP $\,a = 2,\, n = 100\,$ yields $\,2^{100}\,(0,\ 3)\,=\, 3\cdot 2^{100}$

and the 2nd: $\, \ a = 3,\, n = 101\,$ yields $\, 3^{101}(-2,4) = 2\cdot 3^{101}$

Both are, in fact, correct. Notwithstanding, the result becomes obvious and less convoluted if you consider the well known identity$$\text{gcd}(a,b)=\text{gcd}(b-a, b)$$