The answer to the first question is "no". You can see this with specific examples, but here is a more conceptual approach: Take $Y$ an uncomplemented subspace of $X$ and in $Z:= X\oplus_1 X$ identify $Y\oplus 0$ with $0 \oplus Y$ in the obvious way; that is, mod out from $Z$ the subspace $\{(y,-y) | y \in Y\}$. $X\oplus 0$ and $0 \oplus X$ are norm one complemented in the resulting quotient space of $Z$ but their intersection $Y \oplus 0 = 0 \oplus Y$ is not complemented. (This is just a categorical push out construction specialized to the appropriate category of Banach spaces.)

The answer is yes if the subspaces are norm one complemented and the space $X$ is uniformly convex. This is intuitive, because if $P$ is a norm one projection on a uniformly convex space and $x$ is not in the range of $P$, then $\|Px\| < \|x\|$, since otherwise all vectors on the line segment from $x$ to $Px$ would have norm $\|x\|$. Hence one guesses that playing ping pong with two norm one projections $P$ and $Q$ will produce a norm one projection onto $PX\cap QX$. To see that this works without doing any computations or calculating rates of convergence (at the risk of making experts cringe), set $P_1=P$, $P_{2n} = QP_{2n-1}$, $P_{2n+1}=PP_{2n}$. Let $x\in X$ and let $a=a(x)$ be the limit of the nonincreasing sequence $\|P_n x\|$. I claim that $\|P_{n+1}x - P_{n}x\| \to 0$. Indeed, (1/2)$\|P_{n+1}x + P_{n}x\|$ also converges to $a$, so the claim follows from the uniform convexity of $X$. Let $V$ be a limit in the weak operator topology of some subnet of $P_{2n}$. By the claim, the corresponding subnet of $P_{2n+1}$ also converges to $V$ in the weak operator topology. From this it is evident that $V$ is a norm one projection onto $PX\cap QX$.

ADDED 13 Jan. 2012: Notice that in the first construction $X$ can be uniformly convex, in which case $Z$ (and therefore also every quotient of $Z$) is isomorphic to a uniformly convex space.