Let $f(t)=t^2$. Find the best approximation to $f(t)$ over $[0,1]$.
Consider the Banach space $(C[0,1],\|\cdot\|_\infty)$. Then the best constant approximation, in my understanding, should be $g(t)\equiv 1$, since
$$\left\|t^2\right\|_\infty=1$$
In $\left(L^1[0,1], d_1\right)$ we have
$$\int_0^1|t^2|dt=\int_0^1t^2dt=\frac{1}{3}$$
So the best approximation in this case is $g(t) \equiv \frac13$.
In $\left(L^2[0,1], d_2\right)$ we have
$$\left(\int_0^1t^4dt\right)^{1/2}=\frac{1}{\sqrt{5}}$$
So the best approximation in this case is $g(t) =\frac{1}{\sqrt{5}}$.
Please let me know if my understanding is correct.
Also, it is noted that the result for $L^2[0,1]$ can be obtain in at least two ways. What could a second way possibly be? I'd appreciate a hint on this.
In all three cases, you found the norm of the function $f(t)=t^2$ in the respective space instead of addressing the actual question. But the norm $\|f\|$ is not necessarily the best constant approximation to the function $f$. As was already pointed out in the comments, you need to minimize $\|f(t)-c\|$ in each case.
Let's consider $C[0,1]$ with the supremum norm $\|h\|_{\infty}=\sup\limits_{t\in[0,1]}|h(t)|$. You correctly found the norm of $f(t)=t^2$ as $\|f\|_{\infty}=1$, but that wasn't the question. What you need is to find $$\inf_{c\in\mathbb{R}}\|f-c\|_{\infty}=\inf_{c\in\mathbb{R}}\left(\sup_{t\in[0,1]}|f(t)-c|\right).$$
Let's look at some examples first. If $c=0$, then $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|t^2-0|=1$, attained at $t=1$. If $c=1$, then $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|t^2-1|=1$, attained at $t=0$. But both are far from the best (minimal) value.
Note that the values of $f(t)=t^2$ on $t\in[0,1]$ range from $0$ to $1$. Then the triangle inequality tells us that the norm $\|f-c\|_{\infty}$ can't be less than $1/2$. Indeed, if $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|f(t)-c|<\frac{1}{2}$, then $|f(t)-c|<\frac{1}{2}$ for all ${t\in[0,1]}$, and $|f(1)-f(0)|\le|f(1)-c|+|c-f(0)|<\frac{1}{2}+\frac{1}{2}=1$, contradicting the fact that $f(1)-f(0)=1^2-0^2=1$. So we can't do better than $1/2$. But if you think for a moment, you should realize that we can actually achieve $1/2$, i.e. we can find a constant $c$ such that $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|t^2-c|=\frac{1}{2}$. Roughly speaking, the idea is that in the supremum norm the best $c$ lies right in the middle of the range.
With the integral norms $L^1[0,1]$ or $L^2[0,1]$, the exact approach to finding the right $c$ will be different, but at least the underlying idea is the same: find a value of $c$ such that the calculated value of the norm $\|f-c\|$ will be minimal possible (compared to what we'd get with any other $c$).