Limit problem: $\lim_{t\to1} \frac {\sqrt {2t^2-1}\sqrt[3]{4t^3-3t}-1}{t^2-1}$
$$=\frac{\sqrt{2t^2-1}-1}{t^2-1}\sqrt[3]{4t^3-3t}+\frac{\sqrt[3]{4t^3-3t}-1}{t^2-1}$$ now multiply by conjugates and you get the right answer.
As for the second part is possible that the other value of $t$ gives the same value as the limit to 1.
Elegant, I do not know, but a possible solution.
Let $t=x+1$ making the expression to be $$A=\frac{\sqrt{2 x^2+4 x+1} \sqrt[3]{4 x^3+12 x^2+9 x+1}-1}{x (x+2)}$$ Now, using Taylor around $x=0$ or binomial expansion $$\sqrt{2 x^2+4 x+1}=1+2 x-x^2+2 x^3+O\left(x^4\right)$$ $$ \sqrt[3]{4 x^3+12 x^2+9 x+1}=1+3 x-5 x^2+\frac{67 }{3}x^3+O\left(x^4\right)$$ $$\sqrt{2 x^2+4 x+1} \sqrt[3]{4 x^3+12 x^2+9 x+1}=1+5 x+\frac{34 x^3}{3}+O\left(x^4\right)$$ making $$A=\frac{5}{2}-\frac{5 }{4}x+O\left(x^2\right)$$
Concerning the second point, may I suggest you plot thr function for $-1 \leq t \leq 2$ ? You should understand why this number.
You did a hard work, but it should be better to add some justifications when you use $$\lim_{t\to 1}(f(t)+g(t))=\lim_{t\to 1}f(t)+\lim_{t\to 1}g(t)\tag1$$ $$\lim_{t\to 1}f(t)g(t)=\lim_{t\to 1}f(t)\lim_{t\to 1}g(t)\tag2$$ since these do not always hold.
$$\begin{align}\lim_{t \to 1}\frac {\sqrt{2t^2-1}×\sqrt[3]{4t^3-3t}-1}{t^2-1}&=\lim_{t \to 1} \frac {\sqrt[3]{4t^3-3t}-\frac{1}{\sqrt{2t^2-1}}}{\frac{t^2-1}{\sqrt{2t^2-1}}}\\\\&=\lim_{t \to 1}\frac{4t(t^2-1)+t-\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}}{(t^2-1)×\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\end{align}$$
This is correct. (It seems that you used $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$)
$$\begin{align}&\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&\qquad\qquad\quad+\lim_{t \to 1}\frac{\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\frac{4}{3}+\frac{1}{3}\lim_{t \to 1}\frac{t(2t^2-2)×\sqrt{2t^2-1}+t×\sqrt{2t^2-1}-1}{(t^2-1)×(2t^2-1)×\sqrt{2t^2-1}}\end{align}$$
This is correct, but it should be better to add some explanations about why you can use $(1)(2)$ here. I think using $(1)(2)$ without any justifications is not good.
You can avoid using $(1)(2)$.
Using your manipulations, we have
$$\begin{align}&\lim_{t \to 1}\frac{4t+\frac{t(2t^2-1)×\sqrt{2t^2-1}-1}{(2t^2-1)×\sqrt{2t^2-1}×(t^2-1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{t(2t^2-2)×\sqrt{2t^2-1}+t×\sqrt{2t^2-1}-1}{(t^2-1)×(2t^2-1)×\sqrt{2t^2-1}}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{t×\sqrt{2t^2-1}-1}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{2t^4-t^2-1}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{(t^2-1)(2t^2+1)}{(t^2-1)(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\lim_{t \to 1}\frac{4t+\frac{2t}{2t^2-1}+\frac{(2t^2+1)}{(2t^2-1)\sqrt{2t^2-1}×(\sqrt{2t^4-t^2}+1)}}{\left[ \frac{\sqrt[3]{(4t^3-3t)^2}}{\sqrt{2t^2-1}}+\frac{\sqrt[3]{4t^3-3t}}{2t^2-1}+\frac{1}{(2t^2-1)×\sqrt{2t^2-1}}\right]}\\\\&=\frac{4+2+\frac 32}{1+1+1}=\frac 52\end{align}$$