The first one is expressible in terms of the modified Bessel function of the first kind:

$$1+\cfrac1{2+\cfrac1{3+\cfrac1{4+\cdots}}}=\frac{I_0(2)}{I_1(2)}=1.433127426722\dots$$

The second one, through an equivalence transformation, can be converted into the following form:

$$1+\cfrac1{\frac12+\cfrac1{\frac23+\cfrac1{\frac38+\cfrac1{b_4+\cdots}}}}$$

where $b_k=\dfrac{k!!}{(k+1)!!}$ and $k!!$ is a double factorial. By Van Vleck, since

$$\sum_{k=1}^\infty \frac{k!!}{(k+1)!!}$$

diverges, the second continued fraction converges. This CF can be shown to be equal to

$$\frac1{\tfrac1{\sqrt{\tfrac{e\pi}{2}}\mathrm{erfc}\left(\tfrac1{\sqrt 2}\right)}-1}=1.904271233329\dots$$

where $\mathrm{erfc}(z)$ is the complementary error function.

**Establishing the value of the "continued fraction constant" (a short sketch)**

From the modified Bessel differential equation, we can derive the difference equation

$$Z_{n+1}(x)=-\frac{2n}{x}Z_n(x)+Z_{n-1}(x)$$

where $Z_n(x)$ is any of the two solutions $I_n(x)$ or $K_n(x)$. Letting $x=2$, we obtain

$$Z_{n+1}(2)=-n\,Z_n(2)+Z_{n-1}(2)$$

We can divide both sides of the recursion relation with $Z_n(2)$ and rearrange a bit, yielding

$$\frac{Z_n(2)}{Z_{n-1}(2)}=\cfrac1{n+\cfrac{Z_{n+1}(2)}{Z_n(2)}}$$

A similar manipulation can be done in turn for $\dfrac{Z_{n+1}(2)}{Z_n(2)}$; iterating that transformation yields

$$\frac{Z_n(2)}{Z_{n-1}(2)}=\cfrac1{n+\cfrac1{n+1+\cfrac1{n+2+\cdots}}}$$

Now, we don't know if $Z$ is $I$ or $K$; the applicable theorem at this stage is Pincherle's theorem. This states that $Z$ is necessarily the *minimal* solution of the associated difference equation if and only if the continued fraction converges (which it does, by Śleszyński–Pringsheim). Roughly speaking, the minimal solution of a difference equation is the unique solution that "decays" as the index $n$ increases (all the other solutions, meanwhile, are termed *dominant* solutions). From the asymptotics of $I$ and $K$, we find that $I$ is the minimal solution of the difference equation ($K$ and any other linear combination of $I$ and $K$ constitute the dominant solutions). By Pincherle, then, the continued fraction has the value $\dfrac{I_n(2)}{I_{n-1}(2)}$. Taking $n=1$ and reciprocating gives the first CF in the OP.

Here's a short *Mathematica* script for evaluating the "continued fraction constant", which uses the Lentz-Thompson-Barnett method for the evaluation:

```
prec = 50;
y = N[1, prec]; c = y; d = 0; k = 2;
While[True,
c = k + 1/c; d = 1/(k + d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^(-prec), Break[]];
k++];
y
1.4331274267223117583171834557759918204315127679060
```

We can check the agreement with the closed form:

```
y - BesselI[0, 2]/BesselI[1, 2] // InputForm
0``49.70728038020511
```

**Alternative expressions for the second continued fraction**

Just to thoroughly beat the stuffing out of this question, I'll talk about a few other expressions that are equivalent to the OP's second CF.

One can build the Euler-Minding series of the continued fraction:

$$1+\sum_{k=0}^\infty \frac{(-1)^k (k+2)!}{B_k B_{k+1}}$$

where $B_k$ is the denominator of the $k$-th convergent of the continued fraction, which satisfies the difference equation $B_k=B_{k-1}+(k+1)B_{k-2}$, with initial conditions $B_{-1}=0$, $B_0=1$. OEIS has a record of this sequence, but there is no mention of a closed form.

One can also split the original continued fraction into odd and even parts, yielding the following contractions:

$$3-\cfrac{6}{8-\cfrac{20}{12-\cfrac{42}{16-\cdots}}}\qquad \text{(odd part)}$$

$$\cfrac1{1-\cfrac{2}{6-\cfrac{12}{10-\cfrac{30}{14-\cdots}}}}\qquad \text{(even part)}$$

The utility of these two contractions is that they converge twice as fast as the original continued fraction, as well as providing "brackets" for the value of the continued fraction.

**Much, much later:**

Prompted by GEdgar's question, I have found that the second CF does have a nice closed form. Here is a derivation:

The iterated integrals of the complementary error function, $\mathrm{i}^n\mathrm{erfc}(z)$ (see e.g. Abramowitz and Stegun) satisfy the difference equation

$$\mathrm{i}^{n+1}\mathrm{erfc}(z)=-\frac{z}{n+1}\mathrm{i}^n\mathrm{erfc}(z)+\frac1{2(n+1)}\mathrm{i}^{n-1}\mathrm{erfc}(z)$$

with initial conditions $\mathrm{i}^0\mathrm{erfc}(z)=\mathrm{erfc}(z)$ and $\mathrm{i}^{-1}\mathrm{erfc}(z)=\dfrac2{\sqrt\pi}\exp(-z^2)$.

This recurrence can be rearranged:

$$\frac{\mathrm{i}^n\mathrm{erfc}(z)}{\mathrm{i}^{n-1}\mathrm{erfc}(z)}=\frac1{2z+2(n+1)\tfrac{\mathrm{i}^{n+1}\mathrm{erfc}(z)}{\mathrm{i}^n\mathrm{erfc}(z)}}$$

Iterating this transformation yields the continued fraction

$$\frac{\mathrm{i}^n\mathrm{erfc}(z)}{\mathrm{i}^{n-1}\mathrm{erfc}(z)}=\cfrac1{2z+\cfrac{2(n+1)}{2z+\cfrac{2(n+2)}{2z+\dots}}}$$

(As a note, $\mathrm{i}^n\mathrm{erfc}(z)$ can be shown to be the minimal solution of its difference equation; thus, by Pincherle, the CF given above is correct.)

In particular, the case $n=0$ gives

$$\frac{\sqrt\pi}{2}\exp(z^2)\mathrm{erfc}(z)=\cfrac1{2z+\cfrac2{2z+\cfrac4{2z+\cfrac6{2z+\dots}}}}$$

If $z=\dfrac1{\sqrt 2}$, then

$$\frac{\sqrt{e\pi}}{2}\mathrm{erfc}\left(\frac1{\sqrt 2}\right)=\cfrac1{\sqrt 2+\cfrac2{\sqrt 2+\cfrac4{\sqrt 2+\cfrac6{\sqrt 2+\dots}}}}$$

We now perform an *equivalence transformation*. Recall that a general equivalence transformation of a CF

$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cdots}}}$$

with some sequence $\mu_k, k>0$ looks like this:

$$b_0+\cfrac{\mu_1 a_1}{\mu_1 b_1+\cfrac{\mu_1 \mu_2 a_2}{\mu_2 b_2+\cfrac{\mu_2 \mu_3 a_3}{\mu_3 b_3+\cdots}}}$$

If we apply this to the CF earlier with $\mu_k=\dfrac1{\sqrt 2}$, then

$$\sqrt{\frac{e\pi}{2}}\mathrm{erfc}\left(\frac1{\sqrt 2}\right)=\cfrac1{1+\cfrac1{1+\cfrac2{1+\cfrac3{1+\dots}}}}$$

Thus,

$$\frac1{\tfrac1{\sqrt{\tfrac{e\pi}{2}}\mathrm{erfc}\left(\tfrac1{\sqrt 2}\right)}-1}=1+\cfrac2{1+\cfrac3{1+\cfrac4{1+\dots}}}$$