In general, there will be no such fixed points, even when the range of $f$ consists of programs for total computable functions. The reason is that we can easily compute a list of programs for distinct total functions. That is, we can arrange that $T_{f(n)}\neq T_{f(m)}$ whenever $n\neq m$. For example, we could for each $n$ let $f(n)$ be a program that computes the constant value $n$ function, which would certainly achieve $T_{f(n)}\neq T_{f(m)}$. The point about this now is that if $\varphi$ is a computable function such that $\varphi(v)\neq v$ for every $v$, such as the successor function $\varphi(v)=v+1$, then it follows that $T_{f(\varphi(v))}\neq T_{f(v)}$, and so there is no fixed point in your sense.
Edit. François pointed out in the comments that you want $\varphi$ among the $T_{f(n)}$, a feature my particular example above does not have. But this is easily fixed. Let's use $T_{f(n)}(x)=x+n$ instead, and consider $\varphi(v)=v+1$, which is $T_{f(1)}$. The point, as above, is that $\varphi(v)\neq v$ and so also $T_{f(\varphi(v))}\neq T_{f(v)}$. Thus, there is no fixed point.
No, we can construct a computable tree with no computable paths such that there is a probabilistic Turing machine which with nonzero probability constructs a path.
The basic idea is this: kill off a small-measure subset of $2^\omega$ containing all computable reals, and pick our branch completely at random (so that our probability of success is exactly the measure of the complement of the set we kill off).
Here's one way to do it. Call a sequence $\sigma\in 2^{<\omega}$ dangerous if for some $e$, we have $\Phi_e(n)[\vert\sigma\vert]\downarrow=\sigma(n)$ for all $n<4^e$. Note that the set of dangerous strings is computable, and that every infinite computable sequence extends a dangerous string.
Now let $T$ be the tree of all non-dangerous strings, and use the probabilistic Turing machine which simply picks a random extension without thinking at all. Since the function $e\mapsto 4^e$ grows sufficiently fast, we have a positive probability of getting a path. However, $T$ clearly has no computable paths.
Re: question $2$, I don't have an answer but here's a quick observation: we can defeat the coin-flipping algorithm. (I suspect the answer to 2 is yes, by diagonalizing against probabilistic Turing machines - kill nodes to shrink the measure of branches built by a given machine - but I don't immediately see the details.)
Say $\sigma$ is risky if for some $e$, we have
$\Phi_e(2^n)[\vert\sigma\vert]\downarrow=\sigma(2^n)$ for all $n<4^e$, or
for some $m$ not of the form $2^k$ we have $\sigma(m)\not=1$ (say).
Again, take the tree of non-risky strings. Note that every computable sequence extends a risky string, so this tree has no computable paths, and every sequence with only non-risky initial segments has asymptotic-density-$1$ many $1$s, so the set of paths is null.
Best Answer
John Myhill gave an example of a recursive function defined on a compact interval and having a continuous derivative that is not recursive [Michigan Math. J. 18 (1971), 97-98, MR0280373]. However, Pour-El and Richards have shown that if a recursive function defined on a compact interval has a continuous second derivative, then it has a recursive first derivative [Computability and noncomputability in classical analysis, TAMS 275 (1983), 539-560, MR0682717].